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?

• Wiki states that we need to assume this in addition to P=BPP, which makes it trivial. This implication is not known unconditionally. Jun 26 '20 at 17:05

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 $$NPC \cap RP = \emptyset$$.