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. Let me explain this step in more detail. Suppose, for the sake of contradiction, that $h_i$ is a basis for $R$, for $i$ running over some index set $I$. Let $f = \sum_{i \in I} a_i h_i$ and let $g = \sum_{i \in I} b_i h_i$, for $a_i$ and $b_i \in \mathbb{Z}[x]$. Then $2 a_i = x(x-1)(x-2)(x-3) b_i$ for every $i$. In the ring $\mathbb{Z}[x]$, if $2a = x(x-1)(x-2)(x-3) b$ then $2$ divides $b$. Set $b_i = 2 c_i$ for $c_i \in \mathbb{Z}[x]$. Then $\sum c_i h_i$ is an element of $R$ which obeys $2 \sum c_i h_i = g$.
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.
