Assuming we are using the standard definition of a direct product, I think this is never the case.
An integral domain is a commutative ring with unity that has no zero divisors. Just the fact that both $X$ and $R$ have a multiplicative identity element, allows us to pick two elements of this direct product set, specifically $(r, 0_X)$ and $(0_R, x)$ for any $r\in R, x\in X$ and we will always be able to get a zero as a product of two non-zero elements.
Am I correct here, or is there perhaps something that I'm missing?
Note that I did consider the case where both $R$ and $X$ are trivial, but by definition of integral domain, trivial rings cannot be considered as integral domains.
Any and all help is appreciated! Thank you in advance.
A product of rings is an integral domain if and only if exactly one of them is an integral domain and all the others are the zero ring. Otherwise there's indeed a phenomenon like $(1,0)\cdot (0,1)=(0,0)$.
Added: I should point out that my assertion works under the convention that the $0$ is not an integral domain: for, if it were, then the product of the empty family would be the "$0$ domain", but it wouldn't be true that exactly one of the rings in the family is a domain (since exactly zero of them are). So the correct formulation in the (non-standard) case where we allow $0$ to be a a domain would be that a product of rings is "domain-or-zero" if and only if all of them are domains-or-zero and at most one of them is non-zero.