Loaded Q-factor of parallel RLC with series resistive load
Regarding any AC analysis on this circuit, the effect of R and Rload are effectively in parallel and will therefore produce the same Q irrespective of whether their parallel value is in the feed position or the shunt position.

Think about the voltage source and its equivalent circuit with R and Rload - forget about L and C for the moment. Try this for size: -



The two circuits are identical and the little_r resistor has changed to be across a current source. This puts it in parallel with any resistor across terminals A and B.

In fact if V1, Rload (or R1) were inside a box and you were not allowed to look inside, you could never know that what was contained was a voltage source in series with a resistor OR a current source in parallel with a resistor - there's no way of telling.

Here's an even more complex scenario: -



Now, if R2 were zero, the equivalent output impedance is the parallel arrangement of the two other resistors. Does that ring a bell?

It's called Norton's theorum - try googling it - also look up Thevenin's theorum - it operates in the reverse: -



Picture stolen from here. Does this make sense now?

So if you agree that Q = \$\omega C R\$ then you know what value to use for R.



Calculating Q for a parallel LC fed by a voltage source via a resistor. Start with the impedance of a pure LC tuned circuit. This is: -

\$\dfrac{sL}{s^2LC+1}\$ then calculate what Vout would be i.e. the transfer function: -

H(s) = \$\dfrac{\dfrac{sL}{s^2LC+1}}{R + \dfrac{sL}{s^2LC+1}}\$

A bit of algebra and this becomes: -

\$\dfrac{\dfrac{s}{CR}}{s^2+\dfrac{1}{LC} +\dfrac{s}{CR}}\$

The bottom line is clearly (to some LOL) recognizable as the denominator in any damped resonant circuit where the various artefacts are: -



See this document and read the bandpass section to confirm. The above picture is an extract.

So \$2\zeta\omega_0 = \dfrac{1}{CR}\$ and, because Q is \$\dfrac{1}{2\zeta}\$, Q = \$CR\omega_0\$