I am not familiar with playing Russian roulette, neither with revolvers.
In Russian roulette, the chamber is revolved rapidly, but they do not wait for friction to stop the chamber from rotating, this is done manually. So at this stage how large is the friction force compared to the gravitational force due to assymetric distribution of bullets? In this case, I think this additional force is negligible, as the moment of inertia of the revolving part will be way larger than the bullets.
Furthermore, where on one side the bullet accelerates the chamber, it will decelerate at the other side, this will more or less average out, so there will be no preference due to this issue.
Suppose you don't interfere with the revolvers rotation, then when it is almost not rotating anymore, at some point the gravitational force of this bullet will not be negligible any longer. However, at this point there will be mechanical restrictions on the revolver that force the chamber in one of six positions, which will probably prevent this from happening.
Maybe these roulette players point the barrel at this stage in the air or to the ground?