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(*Disclaimer:* I'm neither well-versed in computational complexity nor am I familiar with the relevant literature, so I may very well have missed something.)

By Ladner's theorem, if NP is unequal to P then there are intermediates between NP and P. Thus, if the best NP-complete problem is in QP, then there are problems in NPI.

Your suggestion that, if we can find an NP-complete algorithm in QP, then might be able to prove that P = NP is interesting but I am reasonably sure that it cannot be done with usual techniques. For instance I believe there are oracles $Ω$ relative to which NP is in QP but not in P---thus any technique that would prove that $NP \subseteq QP$ implies NP = P would have to be sophisticated enough that it wouldn't extend to a proof that $NP^Ω \subseteq QP^Ω$ implies $NP^Ω = P^Ω$.