Let $R$ be a local Cohen-Macaulay ring. Let $P$ be a minimal prime ideal of $R$.
Is it true that $\operatorname {depth} R/P=\operatorname {depth} R$ ?
Notice that since $R$ is local Cohen-Macaulay, so $\operatorname {depth}R=\dim R=\mathrm{ht}(P)+\dim R/P =\dim R/P$.
So basically I'm asking: Is $R/P$ again a Cohen-Macaulay ring ?
To expand on Mohan's comment, an example appears in Hartshorne's Ample subvarieties of algebraic varieties, Exercise 5.16 on p. 125. The exercise is worked out in Hartshorne's "Complete intersections in characteristic $p > 0$."
To clarify how this example is relevant, we summarize it here. Note that this example was mentioned before in this MO answer by Hailong Dao.
Example. Let $k$ be an algebraically closed field of characteristic $p > 0$. Consider the monomial subalgebra $$k[tu^3,t^3u,t^4,u^4] \subseteq k[t,u].$$ Hartshorne shows that the kernel $I$ of the surjective map $k[x,y,z,w] \twoheadrightarrow k[tu^3,t^3u,t^4,u^4]$ is generated up to radical by two elements $f$ and $g$. The ring $k[x,y,z,w]/(f,g)$ is Cohen–Macaulay since it is a complete intersection. On the other hand, its reduction (which is the quotient by the unique minimal prime) is $k[tu^3,t^3u,t^4,u^4]$, which is not Cohen–Macaulay (since the system of parameters $t^4,u^4$ is not a regular sequence).
To get a local example, just localize everything at the ideal $(x,y,z,w)$.