Yes, whenever the product is not zero-dimensional there is an uncountable chain of primes (in fact, a chain of cardinality $\mathfrak{c}$ and uncountable cofinality). Let $(A_\alpha)_{\alpha\in I}$ be an infinite collection of rings such that $\prod A_\alpha$ is not zero-dimensional. It follows from Gilmer and Heinzer's Theorem 3.4 that we can partition $I$ into infinitely many sets $I_n$ such that $\prod_{I_n} A_\alpha$ is infinite-dimensional for each $n$. It thus suffices to prove the following: Let $A_n$ be an infinite sequence of rings such that $\dim A_n\geq n$ for each $n$. Then there is an uncountable chain of primes in $A=\prod A_n$.

First, for each $n$ choose a chain of primes $P^n_0\subset P^n_1\subset\dots\subset P^n_n\subset A_n$ for each $n$ and fix a nonprincipal ultrafilter $U$ on $\mathbb{N}$. For any function $f:\mathbb{N}\to\mathbb{N}$ such that $f(n)\leq n$ for all $n$, let $P_f$ be the ideal of sequences $(y_n)\in A$ such that $\{n:y_n\in P^n_{f(n)}\}\in U$. Since $U$ is an ultrafilter, $P_f$ is prime. Write $f\sim g$ if $\{n:f(n)=g(n)\}\in U$ and $f<g$ if $\{n:f(n)<g(n)\}\in U$. Since $U$ is an ultrafilter, $<$ is a total ordering on the set of $\sim$-equivalence classes of functions. Furthermore, whenever $f<g$, $P_f\subset P_g$. The collection of all primes of the form $P_f$ is thus a chain, and is in bijection with the set of $\sim$-equivalence classes of functions $f$ such that $f(n)\leq n$. This latter set, also known as the ultraproduct $\prod_U [n]$ of the sets $[n]=\{0,1\dots,n\}$, is well-known to be countably saturated and have cardinality $\mathfrak{c}$.