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Consider $A=\mathbf{Z}_p[x,y]/(xy-p)$ and $\bar I = (x,y)$. Suppose $I$ lifts $\bar I$ and does not contain $p$; let $I'=(I,p^2)$, which has the same property as $I$. Then the $\mathbf{Z}/p^2$-algebra $B=A/I'$ satisfies $B/p\cong \mathbf{F}_p$, $p^2 B = 0$ and $pB\neq 0$, so we have $B\cong \mathbf{Z}/p^2\mathbf{Z}$. The map $A\to B=\mathbf{Z}/p^2\mathbf{Z}$ furnishes a solution $(x,y)\in (\mathbf{Z}/p^2\mathbf{Z})^2$ of the equation $xy=p$, but there are none. Contradiction.