Would xoring 2 independant AES CTRs to produce p-rand introduce vulnerabilities?
No, this is safe.

In fact, if you show a way of distinguish the stream $AES_{k_1}(C) \oplus AES_{k_2}(C)$ from a random stream with fewer than $2^{64}$ outputs, you have just demonstrated a way of distinguishing AES from a random permutation.

Here is how this works: suppose we are given Oracle assess to a permutation $P$, which might be $AES_{k_1}$ for some unknown key $k_1$, or might be a truly random permutation.

Then, we ask the Oracle for the values $P(C), P(C+1), ..., P(C+n)$, select a random $k_2$, and compute the sequence $P(C)\oplus AES_{k_2}(C), P(C+1)\oplus AES_{k_2}(C+1), ..., P(C+n)\oplus AES_{k_2}(C+n)$.  Then, we run our stream distinguisher on that.

If $P = AES_{k_1}$ for some $k_1$, then this is precisely your stream, and the distinguisher will say "Yes"

If $P$ is a random permutation, well, with fewer than $2^{64}$ outputs, we can treat it as a random function (as collisions are inprobable).   With $P$ as a random function, any value $(C+i)\oplus AES_{k_2}(C+i)$ is a random value; hence the whole stream is a random string, and so the distinguisher will say "No".