Yes it is. With a small but important caveat.
And the reason is that the key entropy (256 bits) exceeds any possible message entropy (128 bits). Simples.
It's easy to read "AES" and infer that it can't be as good as a one time pad (OTP). But that's a fallacy within this narrow context. Consider correctly implemented AES as a compressive mixing function, say $\theta$. 384 (as 256 + 128) bits go in, 128 bits come out.
$$ AES_k(m) \rightarrow \theta_k \; (m) \rightarrow k \oplus m $$
If the Kolmogorov entropy of $k$ exceeds that of $m$, the system by definition must be a OTP. Recall that the mixing function only became a mathematical XOR when computers arose. Before then all that was required was a keyed bijective relationship between plain text and cipher text. Something like a DIANA table. That works on A-Z letters, and $\theta$ works on 128 bit binary data.
And now that Kolmogorov has been dragged up, to the caveat. $k$ must be truly random. So no counters, random number generators e.t.c. No books on your shelf or music files on your PC. There must be a physical device producing the key material, with an expected entropy $> \frac{1}{2}$ bits/bit.
Seems a daft way to encrypt :-)
