If you take a secret plaintext $P$ of length $n$ and a single-use key $K$, also of length $n$, which is randomly generated (in the cryptographic sense of the term), the ciphertext $C = P \oplus K$ obtained by xoring them does not reveal anything about $P$ to an adversary who does not know $K$, other than the length of $P$. As far as the adversary is concerned, $P$ could be any of the strings of length $n$.
If you first encode the plaintext in Base64 and usee a key $K'$ of length $n' = \mathsf{length}(\mathsf{Base64}(P))$, the ciphertext $C' = \mathsf{Base64}(P) \oplus K'$ does not reveal anything other than the length of $\mathsf{Base64}(P)$, which is in a one-to-one correspondence with the length of $P$. So xoring the Base64 encoding is just as good as xoring the original message.
Note that it's critical that the key is random. If the key itself is encoded in Base64, i.e. if you publish a ciphertext $\mathsf{Base64}(P) \oplus \mathsf{Base64}(K)$, this will leak quite a bit of information about the message. Xoring a random key works because for each position in the ciphertext, every character is possible for the plaintext, corresponding to every character for the key. But, for example, if a character in $\mathsf{Base64}(P) \oplus \mathsf{Base64}(K)$ has bit 0x40 set (assuming an ASCII-based encoding), there's a $52/64$ chance that the corresponding character in $\mathsf{Base64}(P)$ is one of 0123456789-/, because there's a $52/64$ chance that the corresponding character in $\mathsf{Base64}(P)$ is a letter.
While encrypting the result of Base64 encoding does not leak more information that encrypting the original message, the process of doing the Base64 encoding or decoding can leak information about the message through side channels.
