Can AES-GCM be broken if initialisation vector is known?
However, we are using the same nonce/IV to encrypt different data.


No, don't do that


  if someone indeed gets hold of the plain-text nonce/IV and a number of different encrypted messages (which are encrypted using the same key) can they be able to decrypt the messages successfully?


Actually, it doesn't matter if they know the nonce; if they get two messages encrypted with the same nonce, then they can compute the xor of the two plaintexts.  Depending on the nature of the plaintexts, that is often enough for an attacker to deduce the contents of the original plaintexts.  Given the attacker more than two messages from the same nonce, well, that just makes it easier for him.

In addition, knowing two messages with the same nonce, then can deduce the internal value $H$ to GCM; that'll allow him to generate encrypted messages of his own that would pass validation (hence the authentication piece of GCM goes away).


  We have proposed to make this better by using a different nonce for encrypting each message. However, I am not sure if this will still prevent someone from decrypting the message when they can get hold of the plain-text nonce, even when there is only one message that is encrypted with that nonce. 


That is absolutely better.  It doesn't matter in the slightest if the attacker knows the nonce; that knowledge doesn't allow him to deduce anything about the plaintext.  In practice, we often just prepend the nonce to the ciphertext, as that's the easiest way to get the nonce to the decryptor.  What matters greatly is that you never use the same nonce twice.