Why does capacitance not depend on the material of the plates?
Yes that is true, capacitance is:
\$C = \frac q V\$
where q is the charge and V the voltage between the plates.
As long as the charge \$q\$ can be "hold in place" this relation applies. I mean, there is no need to have a "good" conductor as the charge is static, it does not move.
So as long as for a certain voltage \$V\$ is applied resulting in a certain charge \$q\$ to be present on the capacitor's plates then \$C\$ can be determined.
It does not matter if the plates are bad conductors (high resistance) as it will then simply take longer for all charge to reach its final location. In the final state there will be no difference compared to a capacitor with well conducting plates as the amount of charge will be the same.
Only if you look at the dynamic behavior of a capacitor (how does it respond to quick voltage changes) would you see an influence of the conductivity of the plates. In first order the capacitor would exhibit additional series resistance.
The active part of a capacitor is the dielectric. That's where the energy is stored, that's what the voltage is developed across. The plates just transport current to the right places. A high resistance here could make the capacitor lossy, but will not change the capacitance.
In much the same way, the resistance of a resistor depends on the material and geometry of the resistive part, not the leads.
The active part of an inductor is the iron, ferrite or air-space within the coils, because that's where the energy is stored. High resistance wires will make the inductor lossy, but won't change the inductance.
Typical capacitor plates are made of conductors (metals) which have a huge number of charge carriers. Consider that (very roughly) \$N_A = 6×10^{23}\$, while \$C = 6×10^{18}e\$, so 1 mol of metal has enough charge carriers for 100000 C, assuming one mobile electron per atom. In a capacitor of 1000μF at 100V with Aluminium plates, only 27μg of Aluminium atoms have to donate/accept a single electron to hold the charge, the rest of the atoms stay neutral. Assuming the plates weight 5g, that's 99,9995% of neutral atoms plus 0,0005% of atoms missing one electron. Clearly, a typical capacitor will fail due to breakdown long before the lack of charge carriers in the plates will become apparent.
Things change in semiconductors, where the amount of free carriers is much smaller and depends on the doping. Even then, it's often easier to calculate the capacitance as a static approximation, assuming that the plates stay perfectly conductive and only the distance between them changes as the depletion region grows. It's not always possible though: in fast dynamic processes junction capacitance can only be adequately described using equations for charge flow (e.g. this one), and the solutions indeed depend on the material of the plates.