You are confusing the Q of the inductance with the Q of the circuit. In a series LCR circuit, the Q of the inductor is the ratio of the voltage across the inductor to the voltage across the circuit. As you say, this is also the ratio of the inductive reactance to the resistance. In a circuit with series R & L in parallel with C, this relation also holds.
In a parallel LCR circuit, the resistance is no longer a property of the inductor itself which is why you get a different definition for Q. The Q in this case is more a property of the circuit than the inductor. You cannot convert one circuit to the other although you could find two circuits with identical Q.
The Laplace transforms of the impedance for (r + L) // C is ...
$$\frac{\frac{r}{LC}+\mathrm s\frac{1}{C}}{\mathrm s^2+\mathrm s\frac{r}{L}+\frac{1}{LC}}$$
for which Q=\$\frac{1}{r}\sqrt{\frac{L}{C}}\$ (at resonance)
The Laplace transform of the impedance for R // L // C is ...
$$\frac{\mathrm s\frac{1}{C}}{\mathrm s^2+\mathrm s\frac{1}{RC}+\frac{1}{LC}}$$
for which Q=\$R\sqrt{\frac{C}{L}}\$ (at resonance)
The refection coefficient due an impedance mismatch is: -
\$\dfrac{R-Z_o}{R+Zo}\$
Where Zo is the impedance of the cable and R is the source or load resistance.
And, for your 50/75 ohm setup will be -0.2. So the signal you put down the cable of (say) 3Vp-p will produce a reflection of 0.6Vp-p. Is this too much? It's not great but it's certainly not terrible.
Best Answer
That formula does not apply to distributed inductances.
Coax cable does have a cutoff frequency, when wavelength of the signal gets somewhere on the order of the diameter of the cable (it's some fraction, like 1/4, but I can't remember exactly). For most coax, that cutoff frequency is considerably higher than the useful frequency of the cable because of dielectric dissipation -- it is educational to find specifications for various sorts of coax and look at their attenuation/meter at various frequencies.