Would derandomizing the reduction from SAT to Unique SAT imply $NP$ and $coNP$ are in $\oplus P$?
The standard meaning of "derandomized Valiant-Vazirani theorem" is the following. 


  There exists a deterministic polynomial time algorithm that, given a 3CNF formula $\phi$, outputs formulas $\psi_1, \ldots, \psi_k$ such that
  
  
  If $\phi$ is not satisfiable, then none of the $\psi_i$ are.
  If $\phi$ is satisfiable, at least one of the $\psi_i$ has a unique satisfying assignment.
  


Indeed, if the above is true, $\mathsf{NP} \subseteq \mathsf{P}^{\oplus \mathsf{P}}$ (a  comment by Joro suggests that this is what he actually meant). Since $\mathsf{P}^{\oplus \mathsf{P}}$ is closed under complement, it follows that $\mathsf{coNP} \subseteq \mathsf{P}^{\oplus \mathsf{P}}$ holds as well.

If a derandomized Valiant-Vazirani theorem holds relative to $\oplus P$, i.e. with the 3CNF formulas augmented by $\oplus P$ predicates, then, using Fortnow's argument in his simplified proof of Toda's theorem, we would get $\mathsf{PH} \subseteq \mathsf{P}^{\oplus \mathsf{P}}$. The usual randomized Valiant-Vazirani theorem implies a randomized version of this: $\mathsf{PH} \subseteq \mathsf{BPP}^{\oplus \mathsf{P}}$. This is one of the lemmas used by Toda.

Note: My original answer had a bug, thanks to Emil Jeřábek for pointing it out.