Once you've XORed two messages with the same secret value, the net result is the same as if you had XORed them with each other without using the secret at all.
Given $plaintext_1$ ⊕ $key$ = $cyphertext_1$ and $plaintext_2$ ⊕ $key$ = $cyphertext_2$, then $cyphertext_1$ ⊕ $cyphertext_2$ == $plaintext_1$ ⊕ $plaintext_2$. Because it's XORed twice with the same key, the double XOR becomes the identity function and the key is simply factored out.
If the attacker learns any bits of the plaintext of either message, they can recover those corresponding bits of plaintext from the other cyphertext message, plus they can recover those bits of the key as well.
So it falls to you to determine if $cyphertext_1$ or $cyphertext_2$ have any knowable information in them. It's completely irrelevant if the plaintext data is ASCII, binary, or EBCDIC. If an attacker can discover or guess what any piece of the data is, it's vulnerable.
This is the classic weakness with the Vernam cypher, and is what enabled the Venona decryption of Soviet secrets. And it's why it's no longer a one-time pad cypher if you use either the plaintext or the key more than one time.