No. E.g., suppose that $n=1$, $m=2$, and the pdf $f$ of each of the $m=2$ iid sample points $X_i:=\mathbf c^{(i)}$ ($i=1,\dots,m$) is given by the formula $f(x)=e^{-x-1}1(x>-1)$ for real $x$. Then, by straightforward calculations, $$p(H)=e^{-1}= 0.367\ldots\ne0.506\ldots=\frac{-2+6 e^2-8 e^3+6 e^4+e^6}{3 e^6}=E\hat p_H.$$

No. E.g., suppose that $n=1$, $m=2$, and the pdf $f$ of each of the $m=2$ iid sample points $X_i:=\mathbf c^{(i)}$ ($i=1,\dots,m$) is given by the formula $f(x)=e^{-x-1}1(x>-1)$ for real $x$. Then, by straightforward calculations, $$p(H)=e^{-1}= 0.367\ldots\ne0.506\ldots=\frac{-2+6 e^2-8 e^3+6 e^4+e^6}{3 e^6}=E\hat p_H.$$

Intuitively, the reason for this is clear: the Voronoi diagram is all about distances, whereas a distribution does not have to care about distances at all.