Is PRF XORed with its key still a PRF? (always)

No.

Let $||$ denote the concatenation. Let $F''$ be a PRF on the domain $\mathbb{M} \times \{0,1\}$. Fix any "special message" $m^* \in \mathbb{M}$. Consider the following construction for $F'$: the key of $F'$ is $k_0 || k_1$, where $k_0$ is a random key for $F''$, and $k_1$ is a random string of the same length. On input $m \in \mathbb{M}$, $F'_{k_0||k_1}(m)$ outputs $F''_{k_0}(m||0) || k_1$ if $m = m^*$, and $F''_{k_0}(m||0) || F''_{k_0}(m||1)$ otherwise.

First, observe that $F'$ is still a PRF. It follows directly from the security of $F''$, and by observing that $k_1$ is truly random and is used only once, on the special input $m = m^*$.

Second, let $F_{k_0||k_1}: m \rightarrow F'_{k_0||k_1}(m) \oplus (k_0 || k_1)$. This is clearly not a PRF, because the second half of $F_{k_0||k_1}(m^*)$ is a string of zeroes (which is very unlikely to happen for a random function).

(I'm glossing over minor details about what we assume on the length of the keys, since it's not hard to handle it, just a bit more tedious).