Turing reducibility implies mapping reducibility
The statement is false. 

Say B is the Halting problem and $A = \overline B$. Then, given oracle to the halting problem we can easily decide its complement.

However it is not true that $A \le_m B$ since $B\in RE$ and $A\in coRE$ but both are undecidable (i.e., if $A \le_m B$ was true, then $B=HP$ is both in $RE$ and $coRE$, that is,  $B\in R$ which is a contradiction).