Is the complement of { www | … } context-free?
Still CFL I believe, with an adaptation of the classical proof.  Here's a sketch.

Consider $L = \{xyz : |x|=|y|=|z| \land (x \neq y \lor y \neq z)\}$, which is the complement of $\{www\}$, with the words of length not $0$ mod $3$ removed. 

Let $L' = \{uv : |u| \equiv_3 |v| \equiv_3 0 \land u_{2|u|/3} \neq v_{|v|/3}\}$.  Clearly, $L'$ is CFL, since you can guess a position $p$ and consider that $u$ ends $p/2$ after that.  We show that $L = L'$.


$L \subseteq L'$: Let $w = xyz \in L$.  Assume there's a $p$ such that $x_p \neq y_p$.  Then write $u$ for the $3p/2$ first characters of $w$, and $v$ for the rest.  Naturally, $u_{2|u|/3} = x_p$.  Now what is $v_{|v|/3}$?  First:  


$$|v|/3 = (|w| - 3p/2)/3 = |w|/3 - p/2.$$

Hence, in $w$, this is position:
$$|u|+|v|/3 = 3p/2 + |w|/3 - p/2 = |w|/3 + p,$$
or, in other words, position $p$ in $y$.  This shows that  $u_{2|u|/3} = x_p \neq y_p = v_{|v|/3}$.

If $y_p \neq z_p$, then let $u$ be the first ${3\over2}(|w|/3 + p)$ characters of $w$, so that $u_{2|u|/3}$ is $y_p$; $v$ is the rest of $w$. Then:
$$|u| + |v|/3 = 2|w|/3 + p$$
hence similarly, $v_{|v|/3} = z_p$.


$L' \subseteq L$: We reverse the previous process.  Let $w = uv \in L'$. Write $p = 2|u|/3$. Then:
$$p+|w|/3 = 2|u|/3+|uv|/3 = |u| + |v|/3.$$
Thus $w_p = u_{2|u|/3} \neq v_{|v|/3} = w_{p + |w|/3}$, and $w \in L$ (since if $w$ is of the form $xxx$, it must hold that $w_p = w_{p+|w|/3}$ for all $p$).