Let $I, J \subset S = k[x_1,\dots,x_n]$ be two monomial ideals and $k$ a field. If every element of $S$ which is $S/J$regular is also $S/I$regular is it true that depth$_S S/I \geq$ depth$_S S/J$ ?

$R = k[a,b,c,d]$, $I= (a, b)\cap (c,d)$ and $J = (ac, bd) = (a, b) \cap (a, d)\cap (c, b)\cap (c, d)$. Then $\mathrm{Ass}R/I \subseteq \mathrm{Ass}R/J$. Hence if $x$ is a regular element of $R/J$ then $x$ is a regular element of $R/I$ but $\mathrm{depth}R/I = 1$ and $\mathrm{depth}R/J = 2$. 


The answer is no without the monomial ideal hypothesis. I do not know whether it is true with that hypothesis. There exists prime ideals $P\subset S$ such that $S/P$ is not CohenMacaulay but, and ideal $J\subset P$ with radical of $J$ equals $P$ and $S/J$ CohenMacaulay. Taking $P=I,J$ in your question, gives examples where the above inequality does not hold. It is easy to construct characteristic $p>0$ examples. For a characteristic zero example, see the review of a paper by Cowsik and Nori, MR0393004. 

