It is not free. Set $f(x) = x(x-1)(x-2)(x-3)/2$.
Claim: $f(x)$ is in $R$.
Proof: We have $$\frac{f(x+N)-f(x)}{N} = \frac{N^3+11 N}{2} + (\mbox{an element of } \mathbb{Z}[x,N]).$$ The fraction $(N^3+11N)/2$ is an integer by checking the two possible parities for $N$, and an element of $\mathbb{Z}[x,N]$ is clearly an integer. $\square$
Let $g(x) = 1$. So $x(x-1)(x-2)(x-3) g = 2 f$. In a free $\mathbb{Z}[x]$ module, this would imply that $2$ divided $g$; since $2$ does not divide $g$, this shows that $R$ is not free.
By the way, this also shows that this is not the direct product of some infinite list of free modules, which I would have considered a more natural guess.