The cardinality of the reduced product is always the same as that of the product, modulo omitting finitely many unusually large $X_i$'s. (Even when $I$ is finite, in which case all $X_i$'s should be omitted.)
Arrange the sets $X_i$ in nondecreasing order of size in a wellordered sequence $(X_\alpha){\alpha<\tau}$. (X_\alpha)_{\alpha<\tau}$. We may assume that $\tau$ is a limit ordinal. Otherwise, the last coordinate $\tau-1$ (like any single coordinate) contributes nothing to the reduced product. Omit $X{\tau-1}$ X_{\tau-1}$ 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 $|\tau|\cdot\kappa$ (see note) 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 $|X/{\sim}| = |X| = \kappa$.
Note: The elements ${\sim}$-equivalent to a given $x \in X$ are obtained by selecting finitely many new values from the sets $X_\alpha-\{x_\alpha\}$ to replace the corresponding value of the sequence $x$. There are at least $\sum_{\alpha<\tau} |X_\alpha-\{x_\alpha\}|$ and no more than $\left(\sum_{\alpha<\tau} |X_\alpha|\right)^{<\omega}$ ways of doing this. Since $\tau$ is infinite and the $X_\alpha$'s all have two or more elements, these two bounds are equal to $\sum_{\alpha<\tau} |X_\alpha| = |\tau|\cdot\kappa$.
