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.