Here's a concrete example, inspired by Jason Starr's comment but of smaller dimension.

Let $A=k[x,y,z]_{(x,y,z)} / (x^2z, xyz)$. Let $p=(x,y)A$. Let $M=A/zA$, thought of as a cyclic $A$-module. Then $M \cong k[x,y]_{(x,y)}$ is a maximal Cohen-Macaulay module over $A$, since $\dim A = \depth M = 2$. We have that $p$ is prime because $A/p \cong k[z]_{(z)}$. We have that $M_p = 0$ because $z \notin p$, so $z$ acts like a unit on $A_p$-modules but kills $M$. Finally, $A_p \cong k(z)[x,y]_{(x,y)} / (x^2, xy)$ has dimension 1 and depth 0.

Here's a concrete example, inspired by Jason Starr's comment but of smaller dimension.

Let $A=k[x,y,z]_{(x,y,z)} / (x^2z, xyz)$. Let $p=(x,y)A$. Let $M=A/zA$, thought of as a cyclic $A$-module. Then $M \cong k[x,y]_{(x,y)}$ is a maximal Cohen-Macaulay module over $A$, since $\dim A = \operatorname{depth} M = 2$. We have that $p$ is prime because $A/p \cong k[z]_{(z)}$. We have that $M_p = 0$ because $z \notin p$, so $z$ acts like a unit on $A_p$-modules but kills $M$. Finally, $A_p \cong k(z)[x,y]_{(x,y)} / (x^2, xy)$ has dimension 1 and depth 0.