Let $E_P=E(R/P)$ and $E_Q=E(R/Q),$ $E(X)$ being the injective envelope of $X.$ If $P,Q$ are prime and $R$ is commutative Noetherian, then $\mathrm{Hom}(E_P,E_Q) \neq 0$ if and only if $P \subseteq Q$ (e.g., Proposition 4.21 of "Injective Modules" by Sharpe and Vamos). So, the question has an affirmative answer in the Noetherian case.
Suppose then that $R$ is not Noetherian, and let $x$ be an element of $E_P$ for which $\mathrm{ann}(x)=P.$ We may choose $x = \bar{1} = 1+P,$ for instance. Let $f:E_P \rightarrow E_Q.$ If $f(\bar{1}) \neq 0,$ then $P$ will annihilate $f(\bar{1}),$ so that $P \subseteq \mathrm{ann}(f(\bar{1})) \subseteq Q,$ which is impossible if $P$ and $Q$ are distinct maximal ideals. Hence, $f$ must be identically $0$ on $R/P.$ Can I conclude from this that $f$ must be the zero map on $E_P$? It seems to me I should be able to decide yes or no, but I am not seeing the way forward.