Here is a counterexample.

Let $S$ be a smooth surface (irreducible). Let $C_1, C_2 \subset S$ be two smooth curves in $S$ which happen to be isomorphic. Let $X$ be the result of glueing $C_1$ and $C_2$ by this isomorphism. The singular locus of $X$ is a curve $C$ which is mapped onto isomorphically by $C_1$ and $C_2$ via the morphism $S \to X$. On the other hand, let $Y$ be the result of glueing two copies $S_1, S_2$ of $S$ where $C_1 \subset S_1$ is glued to $C_2 \subset S_2$ via the given isomorphism of $C_1$ with $C_2$. Again the singular locus of $Y$ is a curve $C' \subset Y$ which is mapped onto isomorphically by $C_1 \subset S_1$ and $C_2 \subset S_2$ via the morphism $S_1 \amalg S_2 \to Y$. We may and do think of $S_1$ and $S_2$ as closed subschemes of $Y$ (in fact $S_1$ and $S_2$ are the irreducible components of $Y$). There is a morphism $$ f : Y \longrightarrow X $$ compatible with the obvious morphism $S_1 \amalg S_2 \to S$ and the morphisms $S \to X$ and $S_1 \amalg S_2 \to Y$ mentioned above. For any closed point $c' \in C'$ corresponding to $c = f(c') \in C$ the morphism $f$ is etale at $c'$.

OK, now we are going to choose a curve $D \subset X$ which is smooth, meets $C \subset X$ exactly at $c$. Just take a general smooth curve on $S$ passing through one of the two points above $c$ and take the image. As primes in $\mathcal{O}_{X, c}$ we will use $(0)$ and the prime ideal cutting out $D$. We can do this because $X$ and $D$ are irreducible. Then $f^{-1}(D)$ will have two irreducible components $D_1$ and $D_2$ with $D_1 \subset S_1$ and $D_2 \subset S_2$. But only one of these will pass through $c'$ by our choice of D (to see this I suggest drawing a picture). Say $c' \in D_1$. Then we pick our prime ideal in $\mathcal{O}_{Y, c'}$ to be the prime ideal $\mathfrak P$ cutting out $S_2$ which does indeed lie over $(0) \subset \mathcal{O}_{X, c}$ as $S_2 \to X$ is dominant. But there is no prime containing $\mathfrak P$ in $\mathcal{O}_{Y, c'}$ lying over the prime ideal cutting out $D$ because $D_2$ does not pass through $c'$.

Here is a counterexample.

Let $S$ be a smooth surface (irreducible). Let $C_1, C_2 \subset S$ be two disjoint smooth curves in $S$ which happen to be isomorphic. Let $X$ be the result of glueing $C_1$ and $C_2$ by this isomorphism. The singular locus of $X$ is a curve $C$ which is mapped onto isomorphically by $C_1$ and $C_2$ via the morphism $S \to X$. On the other hand, let $Y$ be the result of glueing two copies $S_1, S_2$ of $S$ where $C_1 \subset S_1$ is glued to $C_2 \subset S_2$ via the given isomorphism of $C_1$ with $C_2$. Again the singular locus of $Y$ is a curve $C' \subset Y$ which is mapped onto isomorphically by $C_1 \subset S_1$ and $C_2 \subset S_2$ via the morphism $S_1 \amalg S_2 \to Y$. We may and do think of $S_1$ and $S_2$ as closed subschemes of $Y$ (in fact $S_1$ and $S_2$ are the irreducible components of $Y$). There is a morphism $$ f : Y \longrightarrow X $$ compatible with the obvious morphism $S_1 \amalg S_2 \to S$ and the morphisms $S \to X$ and $S_1 \amalg S_2 \to Y$ mentioned above. For any closed point $c' \in C'$ corresponding to $c = f(c') \in C$ the morphism $f$ is etale at $c'$.

OK, now we are going to choose a curve $D \subset X$ which is smooth, meets $C \subset X$ exactly at $c$. Just take a general smooth curve on $S$ passing through one of the two points above $c$ and take the image. As primes in $\mathcal{O}_{X, c}$ we will use $(0)$ and the prime ideal cutting out $D$. We can do this because $X$ and $D$ are irreducible. Then $f^{-1}(D)$ will have two irreducible components $D_1$ and $D_2$ with $D_1 \subset S_1$ and $D_2 \subset S_2$. But only one of these will pass through $c'$ by our choice of D (to see this I suggest drawing a picture). Say $c' \in D_1$. Then we pick our prime ideal in $\mathcal{O}_{Y, c'}$ to be the prime ideal $\mathfrak P$ cutting out $S_2$ which does indeed lie over $(0) \subset \mathcal{O}_{X, c}$ as $S_2 \to X$ is dominant. But there is no prime containing $\mathfrak P$ in $\mathcal{O}_{Y, c'}$ lying over the prime ideal cutting out $D$ because $D_2$ does not pass through $c'$.