No.  Fix $k \in \mathbb{N}$ and let $G$ be a graph of treewidth at most $k$. If you analyze their polynomial-time algorithm that decides if $G$ has an equitable tree-colouring, you'll notice that the degree of the polynomial depends on $k$.  For example, in the proof of Theorem 11, there is a bound where $k$ appears in the exponent.  Thus, it is not an FPT algorithm.  Of course, it is widely believed (but not proved) that FPT $\neq$ W[1].