How secure is XOR-CBC?
Completely insecure. 

If you just replace the encryption with XOR with the key, then the first ciphertext block is just $c_1=m_1 \oplus k$. The second block would be $c_2=m_2 \oplus c_1 \oplus k = m_2 \oplus m_1$.

The second block is actually not encrypted at all. And that goes on, the first, third, fifth and so on block is just an XOR with the key, the second, fourth, sixth and so on doesn't depend on the key at all.