Yes, there is such a characterization: these are exactly those functions which are real on the real line.
Define the following operator on the set of entire functions: $f^*=\overline{f(\bar{z})}$. Since $f^{**}=f^*$, your inequality for all $z$ implies equality: $\Re f^*=\Re f$. Thus $f^*=f+ic$, where $c$ is a real constant, and plugging some real $z$ we conclude that $c=0$.