Can AES GCM mode detect an incorrect key and refuse to decrypt?
An AES-GCM-encrypted message is obtained per an encryption key $K_0$. As for any correct implementation of AES-GCM, we assume that legitimate encryption of multiple ciphertexts with $K_0$ uses $\text{IV}$ such that the counters (obtained by increment of the $\text{IV}$) do not overlap; and that the implementation of decryption gives an error message when the authentication check fails.

will AES in GCM mode detect that I am using the wrong key and give me an error message (..) ?

No, with strictly only what's stated. It is possible to exhibit a ciphertext (including $\text{IV}$) and two distinct keys $K_0$ and $K_1$ that both decipher successfully (and a plaintext that, encrypted with key $K_0$ and $\text{IV}$, encrypts to the ciphertext including $\text{IV}$ and authenticator tag, thus matching the wording in the above re-statement of the question).
But yes, if the ciphertext was prepared legitimately (which was my assumption in a first version of this answer). An attack against a given ciphertext, aiming at exhibiting $K_1$, has cost $p\,2^k$ AES encryption for probability $p$ of success and $k$-bit authentication tag, even if multiple ciphertexts are available. Prepared legitimately holds at least if one (or both) of the following holds:

The key $K_0$ is chosen uniformly at random independently of $\text{IV}$ and the plaintext (even if the plaintext, $\text{IV}$, and $K_1$, are chosen maliciously). That includes any case where $K_0$ is kept secret from adversaries at least until production of the plaintext, which is a standard assumption.
The $\text{IV}$ is chosen uniformly at random independently of $K_0$ and the plaintext (as is the case in some but not all legitimate uses of AES-GCM encryption); that's even if the plaintext, $K_0$, and $K_1$, are chosen maliciously.

Argument: from the structure of GCM it can be shown, under a model of AES as an ideal cipher, that the AES-GCM authenticator tag is a PRF of the key and $\text{IV}$, for any fixed message (plaintext, or ciphertext excluding $\text{IV}$ and authenticator tag).

Conclusion: AES-GCM encryption with secret key $K_0$, or with random unpredictable $\text{IV}$, makes it infeasible that later (including if $K_0$ leaks) anyone could find (much less encounter by accident) another key $K_1$ that allows decryption of the unmodified ciphertext (including $\text{IV}$ and authenticator), yielding almost certainly corrupted deciphered plaintext. However, that can be defeated by the party choosing $K_0$/doing the encryption/choosing the message; and the essential unmodified ciphertext hypothesis is artificial, for it goes against the rationale of using authenticated encryption.