Does P not NP imply NP COMPLETE disjoint from RP? According to Wikipedia https://en.wikipedia.org/wiki/RP_(complexity), $P \ne NP$ implies that $RP$ is a strict subset of $NP$. Does anybody have a reference?
Furthermore, am I correct that if this indeed the case, then $NP-COMPLETE \cap RP = \emptyset$ since we can use $NP$ completeness to solve all other $NP$ problems?
You have not accurately summarized the statement in Wikipedia.  The statement in Wikipedia also needs the extra assumption that $P=BPP$, which is widely conjectured but has not been proven to be true.
With that clarification, no reference is needed -- the reasoning is straightforward and already described on Wikipedia.  If $P=BPP$, then $P=RP=co-RP$ (since $BPP=co-BPP$ and $RP \subseteq BPP$ and $co-RP \subseteq BPP$).  If additionally $P \ne NP$, then it follows that $RP \ne NP$.
Yes, if $RP \ne NP$, then $NPC \cap RP = \emptyset$.