Is it true that $FP^{NP[log\cdot n]} = FP^{NP}$ if $P = NP$?
The reverse

The reverse is true. If $FP^{NP[log]} = FP^{NP}$ then $P = NP$. 

You can quickly find this in the Complexity Zoo; It was proven in 
M. Krentel. The complexity of optimization problems, Journal of Computer and System Sciences 36:490-509, 1988.

The actual question

The other way around is easier. To quote the definition from Complexity Zoo:


  $FP^{NP[log]}$: $FP$ With Logarithmically Many Queries To $NP$


Therefore, if $P = NP$, $FP^{NP[log]} = FP^{P[log]}$. Logarithmically many queries to $P$ is certainly a subset of $NP$, so (if $P = NP$) $FP^{NP[log]} = FP^{NP}$.