Why is the voltage in inhomogeneous electric fields undefined?

I presume (perhaps incorrectly) that you are referring to the singular behavior of the electric fields (and potentials) of point charges. The origin of this singularity is a deep question which requires quantum field theory to adequately answer.

Computationally, this issue is brushed under the carpet by setting $V=0$ at infinity, in which case the potential of a point charge a distance $r$ away from the charge is given by $$int_{infty}^r text{d}r ~E(r) = frac{Q}{4pi epsilon_0 r},$$ which is finite for any $r > 0$. However, the singularity you describe still occurs in the $rto 0$ limit. The classical explanation of this points to the fact that the charge distribution of a point charge is given by $$rho(mathbf{r}) = Qdelta^{(3)}(mathbf{r}-mathbf{r}_0),$$ which is infinite as $mathbf{r} to mathbf{r}_0$. In the case of charged parallel plates the charge distribution is "smeared out" and defined by $$rho(mathbf{r}) = sigmadelta(z-z_0).$$ Notice that although the latter also has the delta function, it's a "weaker" singularity in the sense that $$E sim int rho ~ text{d}z = sigma int delta(z-z_0)~text{d}z = sigma quad text{(finite)},$$ but an analogous integral for the point particle will leave a "two dimensional" delta function behind, which exhibits singular behavior.