If L is context-free, must FH(L) be context-free?
The intuition developed in the comments is right. The answer is NO, there is a counter-example, a CFL for which the first halves are not CFL. 

$L = \{ a^m b^n c^n a^{3m} \mid m,n\ge 1 \}$, over the alphabet $\{a,b,c\}$, from the answer on our sister site. 


   Proof by Pumping lemma: pick $a^p b^p c^p \in \mathrm{FH}(L)$; pumping either destroys the "$b^n c^n$"- or the "first half"-property.


A slight adaptation of that language is
$K = \{ a^m b^n c^n \#\# a^{3m} \mid m,n\ge 1 \}$, over the alphabet $\{a,b,c,\#\}$. We can now "force" the point where the cutting site of the first half is and get another proof technique.


   Let $H = FH(K) \cap a^*b^*c^*\#$. This means we only consider first halves where the middle is exactly at the point between the two $\#$-symbols. Thus. $m+2n+1=1+3m$, or $m=n$. Thus $H=\{a^nb^nc^n \mid n\ge 1\}\#$. Now if $K$ is context-free then $H$ is context-free (via the closure property intersection by regular languages). This language is close to a standard non-context-free example $\{a^nb^nc^n \mid n\ge 1\}$. This in turn can be obtained by right quotient with $\#$ which also preserves context-freeness.