How would charge be distributed in charged conductors if the Coulomb law was not ${1}/{r^2}$? Would the excess charge on a conductor move to surface until the electric field inside become zero if the Coulomb law was for example $\frac{1}{r^3}$? If yes, would the distribution $\sigma(x,y)$ be different from when it is $\frac{1}{r^2}$?
 A: 
Would the excess charge on a conductor move to surface untill the electric field inside become zero [...]?

Think about the mechanism of that motion for a moment. The charges moves because


*

*they are free (not bound)

*there exists a non-zero fields, so from $\vec{F}_E = q \vec{E}$ a force on them


Those two facts are independent of the exact form of the Coulomb interactions.  So the short answer is "Yes."

If yes will the distribution $\sigma(x,y)$ be different [...] ?

Sure. WetSavannaAnimal pointed you in the direction of the closed form solution, but it should be intuitively clear that to get the same condition (no field interior to the conductor) with a different form for the field the charge distribution must be different.
Not that the charge distribution may not be strictly a surface distribution at all, and should be written $\rho(\vec{r})$.
A: Suggestion to the question (v3): Generalize the question to a $1/r^s$ potential law in $n$ spatial dimensions! Then according to Henry Cohn's mathoverflow answer here, the charges rush to the boundary iff $s\leq n-2$. So in OP's example $(s=2,n=3)$, the charges don't rush to the boundary, in contrast to the real world $(s=1,n=3)$.
