You could answer this question easily by a simple "back of the enveloppe" calculation.
Assume you have a plate of metal, say iron. Assume you displace all the free electrons from a 1-atom-thickness on one side to the other side. Calculate the charge densities on the surfaces. Calculate the electric field inside the plate.
I expect you to find an enormous electric field. This means that the reaction of a conductor to real-life electric fields should only involve a tiny fraction of the charge available in a thin 1-atom layer of iron. This means that the depletion of free electrons would affect the physics of conductors only in extreme-extreme conditions. Enormous forces would be at play and the material would probably explode.
I expect this result because it is known that the "Coulomb" is an enourmous electrical charge and that it corresponds to the charge of less than a mole of electrons (1E-5 moles actually).
In other words: real life electric fields and voltages are always created by a very small fraction of the available electrons.
Still other words: electric fields in laboratories are always very small compared to electric fields on the atomic scale, really very small.
Why is this so? Simply because materials are are "assembled" by electrical forces. Experiences in electrostatics rarely stress a material very much, specially dense materials. There are exceptions: disruptions in gases,