A word is cubefree if it cannot be written as $xyyyz$ where $y$ has positive length.

Let $h$ be the morphism from $\{0,1,2,3,4\}^*$ to $\{0,1\}^*$ given for words of length 1 as follows ($a\to h(a)$): $$0\to 001001010011$$ $$1\to 001001101011$$ $$2\to 001010011011$$ $$3\to 001101001011$$ $$4\to 010011001011$$ and extending to longer words by the morphism property $h(xy)=h(x)h(y)$.

Is it cubefree-preserving? That is, if $x$ is cubefree then is $h(x)$ cubefree?

(I checked that it is so for $x$ of length at most 8; in general there is no finite test set by a result of Richomme and Wlazinski, but maybe there's something special about this case.)

And if not this map...

does there exist any cubefree-preserving map from an alphabet of size 5 to an alphabet of size 2?