Edited due to mistakes pointed out in the comments:
I think the answer to the problem might be yes. Here are some preliminary thoughts.
First, you know $f_{\nu}(x)$ divides $f(x)$, since $f(x)$ is always in the kernel, the kernel is principal, and $\mathbb{Z}[[x]]$ is a UFD.
Second, you know $f_{\nu}(x)$ is irreducible, else you don't map into a domain.
Third, I think you can probably prove that these polynomials (the $f_{\nu}(x)$) are distinct. If so continue:
Fourth, this all says that $\prod_{\nu}f_{\nu}(x)$ divides $f(x)$ in $\mathbb{Z}[[x]]$. So, to get the result you just have to prove that the constant coefficients agree up to a unit (i.e. up to $\pm 1$). This is equivalent to showing that the constant coefficient of $f_{\nu}(x)$ is $p^{f_{P}\nu_{P}(\alpha)}$ where $P$ is the prime (above $p$) associated to $\nu$ and $f_{P}$ is the inertial degree of $P$ in the ring of integers over $K$. (Sorry for the double use of $f$--for the polynomial and for the inertial degree.)