Does P not NP imply NP COMPLETE disjoint from RP?
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 $NPC \cap RP = \emptyset$.