If a conductor is charged with several megavolts of static electricity, does it have a larger number of free electrons? If so, wouldn't they render the conductor more conductive?
To truly test the conductivity, a generator that was charged to the same potential would need to be employed.
Conductor can't be charged with "static" electricity - any charge that you place on the conductor is a free excessive charge. This charge will rearrange in such a way, that the conductor is equipotential. At the end of this rearrangement, all the excessive charge will be present on the outer shell of the conductor and there will be no electric field inside the conductor. Once you provide a path for current to flow, the conductor will get rid of all the excessive charge.
If you connect a "generator that was charged to the same potential" to this conductor, there will be no electrical current since there are no potential differences. This means that all the excessive charge will remain on the conductor. However, when you'll apply additional DC voltage for resistance measurement, the current will flow through the conductor as though it is completely neutral (remember, there is no electric field inside due to excessive charge).
If you connect a measurement equipment having different potential, the charge on the conductor will again rearrange in such a way that the conductor and the test equipment are equipotential. Usually the conductor will be discharged through the probes. Note: if there is really huge amount of charge on the conductor, the initial current may be so large that the test equipment will get damaged.
Seems like you should not make any special preparations in order to measure the resistivity of the conductor as long as your test equipment have definite potential, and the amount of excessive charge can't damage your test equipment.
However, sometimes you may want to measure the resistivity with a floating meter (any multimeter is floating). What happens in this case? Well, the excessive charge that you placed on the conductor will once again rearrange. It is difficult to predict what will be the final configuration, but you may be 100% sure that there will be no electric field inside the conductor (except for the field induced by the multimeter itself). If the charge re-configuration did not damage your multimeter, you will measure an accurate resistance.
In summary:
If you want to measure DC resistance, the amount of initial charge on the conductor is not important, as long as the discharge of this charge does not damage the test equipment and is not changing the conductor itself (by heating it, for example).
Disclaimer:
I'm 100% sure that for very high charge densities there are exotic quantum-mechanical effects appear (this is always the case - take something to an extreme, and there will be unpredicted complications). I have no clue what these effects may look like and whether they can change the above description, but I guess this stuff is handled by physicists, not engineers.
Back-of-the-envelope fun:
According to http://en.wikipedia.org/wiki/Capacitance#Self-capacitance, a 20-cm radius sphere has a capacitance of ~22 pF. If charged to -1 MV, it would have 22 uC of charge. A coulomb is about 6E18 electrons, so there would be about 1E14 extra electrons on the surface. The area is about half a square meter, or 5E17 square nanometers, and each square nanometer would have (very approximately) 10 or so metal atoms exposed.
So, even accounting for the fact that all the extra electrons would go to the surface of the conductor, even really high voltages would add on average a tiny fraction (about 1/50,000) of an electron per atom to the "electron soup" of the conductor.
I'm not a physicist, but I wouldn't expect that to make much difference to the conductivity of the system.
