There are all sorts of models of $ZFC$ in which every set is definable without parameters, including nonmeasurable sets; indeed a recent paper of Hamkins, Linetsky, and Reitz is devoted to such "pointwise definable" models.
Also, as pointed out in Theo Buehler's comment to the question, there certainly exist definable subsets of reals that are $ZFC$-provably not Borel.
However, the situation is completely different for measurability, thanks to a key result of Harvey Friedman, who proved here that there is a model of $ZFC$ in which there is no definable subset of reals that is not Lebesgue measurable.
Indeed Friedman proved the stronger result that the following property holds in his model:
(*) Every equivalence class of sets of reals modulo null sets that is in $OD(\Bbb{R})$ consists of Lebesgue measurable sets.
Recall that $X$ is in $OD(\Bbb{R})$ if $X$ is definable with parameters from $Ord \cup \Bbb{R}$.
Note that (*) implies that no non-measurable subset of reals in definable, since if $X$ is any definable subset of reals that is not measurable, then the equivalence class $\[X\]$ of $X$ modulo null sets satisfies the following two properties:
(1) $\[X\]$ definable,
(2) No member of $\[X\]$ is measurable.
So, to sum-up, Friedman's theorem provides a negative answer to your question for Lebesgue measurability.