The answer is no in general.
Take $R = \prod_{n \in \mathbb{N}_{> 0}} \mathbb{Z}/2^n\mathbb{Z}$. Then the Jacobson radical of $R$ is $\prod_{n \in \mathbb{N}_{> 0}} 2\mathbb{Z}/2^n\mathbb{Z}$, and it contains a non-nilpotent element, namely $(2 + 2^n \mathbb{Z})_n$. Therefore the Jacobson radical of $R$ doesn't coincide with the nilradical of $R$.
All credits go to Georges Elancwajg, see this MO post.
Side note. Let $J(R)$ denote the Jacobson ideal of $R$. Knowing that $x \in J(R)$ holds if and only if $1 + rx \in R^{\times}$ for every $r \in R$, the identity $J(\prod_{i \in I} A_i) = \prod_{i \in I} J(A_i)$ is immediate.