According to my limited knowledge we know that since Integer Factorization lies in the intersection of NP and co-NP it cannot be NP-complete unless NP=co-NP. However, do we know any other implications yet- like if P != NP, does this mean that Integer Factorization is necessarily not in P? Conversely, if Integer Factorization is in P, does this imply anything about the polynomial hierarchy? Thanks.
Before answering your questions a distinction has to be made, when talking about $NP$ completeness we are talking about the decision problem not the function problem, which is given an integer N and an integer M with 1 ≤ M ≤ N, does N have a factor d with 1 < d < M.
So answering your questions in order, Since integer factorization is not known to be NP complete, $P \neq NP$ would not cause it to automatically not be in $P$ since quite often things that were thought to be in $NP$ are moved to $P$, only an $NP$ complete problem would be excluded from being in $P$. For the second one, again since it is not $NP$ complete a polynomial time solution to it will not effect the hierarchy for the same reason as before.
As an extra note, the actual location of factorization in the hierarchy is unknown, we know is is $BQP$, $NP$, $co-NP$, $FNP$ and $UP$. It may end up being $NP$-Intermediate, but no one knows for sure.