**Edit.** I had posted this answer to complement Eric's original answer, which showed that the number of classes was at least ${\frak c}^+$, since at that time we didn't quite yet know whether there were $2^{\frak c}$ classes. Afterwards, however, Eric improved his answer to get $2^{\frak c}$ directly. Following the comments, though, I have left this answer up.

Let me complement Eric's answer by showing that it is relatively
consistent to have strictly more than ${\frak c}^+$ many
equivalence classes. Indeed, it is relatively consistent with ZFC
to have $2^{\frak c}$ many equivalence class, in a case where this is
larger than ${\frak c}^+$.

Specifically, I claim that if the continuum hypothesis holds and
there is a thick Kurepa tree (an $\omega_1$ tree with
$2^{\omega_1}$ many branches), then there are
$2^{\omega_1}=2^{\frak c}$ many equivalence classes. Indeed, I shall construct an almost-disjoint family of $2^{\omega_1}$ many Vitali sets.

To see this, let $T$ be a thick Kurepa tree, and let $\langle
A_\alpha\mid\alpha<\omega_1\rangle$ enumerate the equivalence
classes of reals under translation-by-a-rational. Label the
$\alpha^{th}$ level of $T$ with the countably many elements of
$A_\alpha$. For any path $s$ through $T$, the set $A_s$ of labels
appearing on the nodes of $s$ will be a Vitali set, and therefore
non-measurable. Further, any two distinct paths $s\neq t$ will
have $A_s\cap A_t$ being countable, and so $A_s\not\sim A_t$.
Since $T$ is a thick Kurepa tree, we therefore have $2^{\omega_1}$
many branches and thus this many equivalence classes modulo your
relation. The collection $\{\ A_s\mid s\in[T]\ \}$ is an almost-disjoint family of $2^{\omega_1}$ many Vitali sets.

Finally, let me explain that it is relatively consistent from an
inaccessible cardinal that there is a thick Kurepa tree, yet CH
holds and $2^{\omega_1}$ is very large. One way to do this is as
follows. Start with $\kappa$ inaccessible in $V$ and $2^\kappa$
very large (by forcing if necessary). Let $V[G]$ be the forcing
extension by the Levy collapse, so that $\kappa=\omega_1^{V[G]}$.
Consider the tree $T=(2^{<\kappa})^V$ in the model $V[G]$. Since
every ordinal less than $\kappa$ was made countable, this has
become an $\omega_1$-tree. Yet, since $2^\kappa$ was very large
and cardinals $\kappa$ and above were preserved, we have
$(2^\kappa)^V$ many branches through this tree. So it is thick.

reallya question about the axiom of choice. I mean, if this is a question about the axiom of choice, we need to start tagging many other questions like that as well. $\endgroup$ – Asaf Karagila Mar 19 '15 at 20:53