If the IV is send with the ciphertext rather than derived (this seems assumed, otherwise the IV is unlikely to get corrupted) and is indeed fully unpredictable the those bits that are affected cannot be decrypted (presuming that the bits are changed with a 50% chance). This is because decryption of the first block of CBC depends on a XOR with the IV to derive the plaintext. If the value of any of the IV bits is unknown, then the value of the plaintext is unknown as well, similar to the way a one-time-pad works.
However, the decryption of the other, unaffected bits is still fully possible. And any block after that only relies on the ciphertext block before it, rather than the IV. So most of the message would still be recoverable, regardless of what happens to the IV.
So basically, the answer is no this is not correct other than for the bits in the first block of the ciphertext / plaintext that are affected by the changes in the IV.