Would $\sf RP = NP$ imply $\sf NP = coNP$?
If we will able prove that RP is closed under complement then we can say that
If RP = NP then it imply NP = Co-NP.

But we don't know whether RP=Co-RP or not.  BPP = P can be proved under some reasonable assumptions but RP $ \subseteq $ BPP. 

If we show that RP = BPP then your statement will be correct.

References:


RP in Wikipedia
Notes on Randomized Algorithms (pdf)
RP in the Complexity Zoo