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The answers currently posted are ignoring a few important details so I’m going to give my own.
I may rehash some things already said.
To make everything absolutely clear I write here a complete derivation of the forced damped oscillator with emphasis on the role of the $Q$ factor.

Basic equations

Consider the equation of motion of a forced, damped harmonic oscillator:

$$\ddot{\phi}(t)+2\beta\dot{\phi}(t) + \omega_0^2 \phi(t) = j(t) \,.$$

Here $\beta$ is a coefficient of friction (for the case where the friction force is proportional to the velocity $\dot{\phi}$), $j$ is an external forcing function, and $\omega_0$ is the un-damped frequency of the system.

We define the fourier transform $\tilde{\phi}(\omega)$ by the equation

$$\phi(t) = \frac{1}{2\pi}\int \tilde{\phi}(\omega)e^{-i \omega t}\,d\omega\,.$$

Plugging the Fourier transform into the equation of motion gives

$$\phi(\omega) = \frac{\tilde{j}(\omega)}{-\omega^2 – 2i\beta \omega + \omega_0^2}
= \frac{\tilde{j}(\omega)}{(\omega-\omega_+)(\omega – \omega_-)}$$

where $\tilde{j}$ is the Fourier transform of $j$ and

$$\omega_{\pm}\equiv -i\beta \pm \omega_0′ \qquad
\omega_0’\equiv \omega_0\sqrt{1-\left( \frac{\beta}{\omega_0} \right)^2}\,.$$

My $\omega_0’$ is what you called $\omega_r$ if we set $\gamma / \sqrt{2} = \beta$.
Note that for light damping, i.e. the case $\beta \ll \omega_0$, we get $\omega_0′ \approx \omega_0$.

In order to understand the meanings of the quality factor $Q$ we investigate these equations for two cases.

Free oscillation

First we consider the case where the forcing function is just an instantaneous whack at $t=0$.
This should cause the system to oscillate but with a decreasing amplitude as energy is lost to friction.
Mathematically we denote the instantaneous whack as $j(t) = A \delta(t)$.
This gives $\tilde{j}(\omega) = A$.
We find $\phi(t)$ by inverse Fourier transform

$$
\begin{align}
\phi(t)
&= \frac{1}{2\pi} \int \tilde{\phi}(\omega)e^{-i\omega t} \, d\omega \\
&= \frac{1}{2\pi} \int \frac{A e^{-i \omega t}}{(\omega-\omega_-)(\omega-\omega_+)} \, d\omega \\
&= \frac{A}{\omega_0′} e^{-\beta t} \sin(\omega_0′ t) \, . \qquad(*)
\end{align}
$$
Let’s understand this result.
At $t=0$ the system is at $\phi=0$.
This makes sense because we whack it at $t=0$ but it hasn’t had time to go anywhere yet.
As time goes on, it oscillates at frequency $\omega_0’$ but with decreasing amplitude.
Note that the oscillation frequency in this case is not the un-damped frequency $\omega_0$; it is $\omega_0’$ which is slightly shifted because of the friction.
Larger friction causes a bigger shift in this free oscillation frequency.
Of course, as you can see, larger friction also causes the amplitude to decrease faster, which makes sense.

Now, what about $Q$?
Suppose $\phi$ represents the position of a mass on a spring.
In that case, the potential energy of the system is proportional to $\phi^2$.
Similarly, if $\phi$ represents the current in an LRC circuit then the the inductive energy is proportional to $\phi^2$.
Twice per oscillation all of the system’s energy goes into the potential (or inductive) energy.
From Eq. $(*)$ that energy is
$$E(t) = E(0) e^{-2\beta t} \,.$$
Now we can easily find $Q$

$$Q \equiv \frac{\text{energy stored}}{\text{energy lost per radian}} =
\frac{E(t)}{-\frac{dE}{dt} \frac{dt}{d\,\text{radians}}}
= \frac{E(0)e^{-2\beta t}}{2\beta E(0) e^{-2\beta t}/\omega_0′}
= \frac{\omega_0′}{2\beta} \,.$$

This is almost exactly the same as your expression $\left( \omega_0^2 – \gamma^2 / 2 \right)^{1/2} / \gamma$ except that I think you have messed up a factor of $\sqrt{2}$ somewhere.
Anyway, the point is that your “more exact” formula for the $Q$ value is really just the $Q$ you get if you consider the case of free oscillation of the damped system.

Steady state driven system

Now let’s consider the case where the system is subjected to constant driving of the form

$$j(t) = A \cos(\Omega t) \, .$$

Then

$$\tilde{j}(\omega) = \frac{(2\pi)A}{2}(\delta(\omega – \Omega) + \delta(\omega + \Omega))\,.$$
Plugging that into the integral and cranking away gives us

$$\phi(t) = \text{Re} \left[ e^{-i\Omega t} \frac{-A}{\Omega^2 + 2i\beta \Omega – \omega_0^2}\right]\, .$$

Let’s look at the case where we’re driving at the natural resonance frequency, i.e. $\Omega = \omega_0$.
In this case we get

$$\phi(t) = \frac{A}{2 \beta \omega_0}\sin(\omega_0 t)\,.$$

The power exerted by the driving force is $\text{force}\times\text{velocity}^{[a]}$:
$$P(t) = j(t)\dot{\phi}(t) = \frac{A^2}{2\beta}\cos(\omega_0 t)^2 = \frac{A^2}{4 \beta}[1 + \cos(2\omega_0 t)] \, .$$
Note that $P(t)$ is always positive.
This is actually the definition of resonance: the resonance frequency is the one such that the work done by the drive is always positive.
In an electrical circuit this is the same as saying that the resonance frequency is the one where the impedance of the damped oscillator is purely real.
No other frequency has this property, so we’ve just shown that $\omega_0$ is the resonance frequency of the damped system$^{[b]}$.
Since we’re in steady state, this work must also be precisely the work lost by the system to the damping.
We can therefore compute the average power loss in one cycle:

$$\langle P_{\text{loss}}\rangle =
\frac{\omega_0}{2\pi} \int_{0}^{2\pi / \omega_0} P(t)\,dt = \frac{A^2}{4\beta}\,.$$

Great.
Now let’s compute the energy stored.
By analogy to the case of a mass on a spring we know that the max potential energy is$^{[c]}$

$$U = \frac{1}{2} \phi_{\text{max}}^2 \omega_0^2 = (1/2)A^2 / 4\beta^2$$

and again since we’re in steady state this is just the total stored energy.
Therefore, the $Q$ value is

$$Q \equiv \frac{\text{energy stored}}{\text{energy loss per radian}}=
\frac{U}{\langle P_{\text{loss}}\rangle / \omega_0}=
\frac{(1/2)A^2 / 4\beta^2}{A^2 / 4 \beta \omega_0} =
\frac{\omega_0}{2 \beta} \, .
$$

This is the expression is almost exactly the same as the one we found for free oscillation, except that now we have $\omega_0$ instead of $\omega_0’$.
Note that we have now answered your question #3, as we have shown that for steady state driving the $Q$ value involves $\omega_0$, not the more complex expression $\omega_0’$.

Answer to the original questions

We’ve seen that we can get two very slightly different expressions for $Q$ depending on whether we consider free oscillation or steady state driving.
In fact, when people talk about $Q$ they’re really talking about the steady state driving one; to keep from getting confused the other expression really shouldn’t be called “$Q$”.
That said, for a system where $Q\gg1$ both expressions give extremely close numbers, so the distinction is mostly academic.

1. Does the energy dissipated per cycle assume that the amplitude is constant from one cycle to the next.

Yes, because when you talk about $Q$ you’re implicitly talking about the steady state driving case in which everything is the same from cycle to cycle.

2. Is it always calculated at the resonance frequency?

By definition, yes. The $Q$ is defined as the energy stored divided by the energy loss per radian in the steady state driving case with drive at the natural oscillation frequency $\omega_0$.

3. If the answer to 2 is yes can you explain why for a forced oscillator system with a damping coefficient of $\gamma$ and natural frequency $\omega_0$ the quality factor is $Q=\omega_0/\gamma$ and not some more complicated expression involving the actual resonant frequency

This was demonstrated in detail in the above discussion/calculation.

Notes:

$[a]$: In fact because of the way the quantities are set up here, if $\phi$ is a displacement of a mass on a spring, then what I’m calling “power” is actually “power divided by mass”.

$[b]$: See this SO question which I posted specifically to help generate this answer.

$[c]$: Again, if you go through and compare to the case of a mass on a spring you’ll see that I’ve left out a factor of the mass.