Would derandomizing the reduction from SAT to Unique SAT imply $NP$ and $coNP$ are in $\oplus P$? The Unambiguous SAT problem (USAT) is to determine whether a 
given formula has a satisfying assignment, when we are 
guaranteed that it has at most one satisfying assignment.

By a theorem Valiant-Vazirani there is probabilistic
reduction from 3SAT to USAT. Given 3SAT formula $\phi$,
over $n$ variables, the algorithm produces $n+2$ formulas
$\phi_i$ with the properties:  if $\phi$ is UNSAT, all $\phi_i$
are UNSAT. Otherwise the probability that $\phi_i$ has exactly
one solution is $\ge \frac18$.

$\oplus P$ is the class of decision problems solvable by a nondeterministic Turing machine in polynomial time, where the acceptance condition is that the number of accepting computation paths is odd.

Since USAT has zero or one solution, $\oplus P$ solves USAT.


  Would derandomizing the reduction SAT to USAT imply $NP$ and $coNP$ are in  $\oplus P$?


This looks plausible to me, though Wikipedia claims 
"there is a relativized universe (see oracle machine) where P = ⊕P ≠ NP = PP = EXPTIME" and $P^{\oplus P}$ is not known to even contain $NP$.

Even it can't be derandomized the reduction to USAT appears
very good in practice given $\oplus P$ oracle.
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.