Is it true that any map from an infinite product of EilenbergMacLane spaces in to an EilenbergMacLane space factor through, upto homotopy, a finite subproduct? Even if we take with field coefficients?

This is not true because it's not true in algebra. Any map $A \to B$ of abelian groups can be realized by a map $K(A,n) \to K(B,n)$ which is unique up to homotopy, and so we can just find a map $\Pi A_i \to B$ which does not factor through a finite subproduct. If $\mathbb F$ is a field, we can construct a map $\Pi^\infty \mathbb F \to \mathbb F$ which does not factor through any finite subproduct using the Axiom of Choice. Let $e_i$ be the element in the product which is 1 in the i'th position and 0 elsewhere. These are linearly independent, and so we can extend them to a basis of $\Pi^\infty \mathbb{F}$. Let $f$ be the unique linear function that sends all the $e_i$ to 1 and sends the rest of the basis elements to 0. Even more interesting is that we can take $\prod_{p \text{ prime}} \mathbb{Z}/p$. Any finite subproduct is a finite group, but this infinite product has nontorsion elements (e.g. $(1,1,1,1\ldots)$ is nontorsion) and so there are nontrivial maps from this infinite product to $\mathbb{Q}$. 

