The cardinality of the reduced product is always the same as that of the product, modulo omitting finitely many unusually large coordinates. (Even when $I$ is finite, in which case all coordinates should be omitted.)
Arrange the sets $X_i$ in nondecreasing order of size in a wellordered sequence $(X_\alpha)_{\alpha<\tau}$. We may assume that $\tau$ is a limit ordinal. Otherwise, the last coordinate (like any single coordinate) contributes nothing to the reduced product. Delete the last coordinate and repeat as long as necessary.
I will show that then $|X| = |X/{\sim}|$ where $X = \prod_{\alpha<\tau} X_\alpha$.
Let $\kappa = \sup_{\alpha<\tau} |X_\alpha|$. Since each ${\sim}$-equivalence class has size
$\sum_{\alpha < \tau} |X_\alpha| = |\tau|\cdot\kappa,$
we have
$|X| \leq |X/{\sim}|\cdot|\tau|\cdot\kappa = \max(|X/{\sim}|,|\tau|,\kappa).$
Since $|X| \geq 2^{|\tau|} > |\tau|$, we conclude that either $|X/{\sim}| = |X|$ or $|X/{\sim}| \leq |X| \leq \kappa$.
Since $\tau$ is a limit ordinal, the diagonal embedding $d:\kappa \to \prod_{\alpha<\tau} |X_\alpha|$, where
$d_\alpha(\xi) = \begin{cases} \xi & when\ \xi < |X_\alpha| \\ 0 & otherwise,\end{cases}$
shows that $\kappa \leq |X/{\sim}|$. So, in the case $|X/{\sim}| \leq |X| \leq \kappa$, we in fact have $|X/{\sim}| = |X| = \kappa$.