Can one claim that AES has perfect secrecy for a key size and message size of 128 bits?
The answer is incorrect, but it's a bit more subtle than it seems. To make this clear, note that encrypting $x$ by computing $c=\operatorname{AES}_{0}(k) \oplus x$ would be perfectly secure (here the key is fixed to 0, but any fixed key value would give the same effect). This is due to the fact that AES is a permutation and so when $k$ is uniformly distributed, then $\operatorname{AES}_0(k)$ is a uniformly distributed value. More formally, for every $c$ and every $x$ there exists a $k$ such that $c=\operatorname{AES}_0(k)\oplus x$. In order to find this, compute $k = \operatorname{AES}_0^{-1}(c \oplus x)$. 

Now, when encrypting $\operatorname{AES}_k(x)$ the same argument fails. This is because for $c=\operatorname{AES}_k(x)$ there is no guarantee that decrypting $c$ under all possible keys gives all possible plaintexts. Thus, it is possible that there is only one $k$ (or a few $k$'s) that give you the known initial portion.