No, if $I$ is uncountable. Indeed, members of
$\mathcal B(\mathbb R)^I$ depend on only countably many coordinates*, but a singleton point is closed in $\mathbb R^I$ and therefore a member of
$\mathbb B(\mathbb R^I)$.
$^*$ proof. The collection of subsets of $\mathbb R^I$ that depend on countably many coordinates is a sigma-algebra, and contains the generating sets
for $\mathcal B(\mathbb R)^I$.
added
I used the usual definition for product sigma-algebra. Given $(X_i, \mathcal F_i)$ for $i \in I$, define the product sigma-algebra $\bigotimes_i \cal F_i$ on the Cartesian product set $\prod_i X_i$ as: The sigma-algebra generated by the sets of the form
$\prod_{i \in I} A_i$,
where $A_i = X_i$ for all but one $i \in I$ and $A_i \in \mathcal F_i$ for the final $i$. This coincides with the notion of "product" from category theory.
Of course, if $I$ is countable, then by countable intersections this coincides with your definition
$$
\sigma \left\{\prod_{i\in I}A_i\mid A_i\in \mathcal F_i, i\in I\right\}
\tag2$$
But if $I$ is uncountable, it is (usually) different. For example, in $\mathbb R^I$, a singleton is not in $\mathcal B(\mathbb R)^I$, even though it is of the form $(2)$. Uncountable intersections are not postulated for a sigma-algebra.
Yes, if $I$ is countable. Then $\mathbb R^I$ is metrizable, and a subbase for the topology is the collection of sets $\prod A_i$ with $A_i = \mathbb R$ for all but one $i$, and $A_i$ open for the final $i$.
Using that, taking finite intersections and countable unions, we see that
all open sets of $\mathbb R^I$ belong to $\mathcal B(\mathbb R)^I$.
