It turns out the answer is "no". There is an example in section 3.4 of Riedtmann's paper "Degenerations for representations of quivers with relations", Ann. sci. Éc. Norm. Sup. v. 19 (1986), 275-301.
The quiver has vertices 1 and 2, with an arrow from 1 to 2 and a loop at 2. The dimension vector is $d=(1,2)$. In $X$, the linear transformation associated to the loop is a single nilpotent Jordan block $N$, and the linear transformation associated to the arrow sends the generator at 1 to an element which is not in the kernel of $N$.
In $Z$, the linear transformation associated to the loop is the same, but the arrow sends the generator at 1 to the kernel of $N$.
It isn't hard to show that $Z$ is in the closure of the orbit of $X$.
$Z$ isn't a direct sum in the way that I asked for because in fact it is indecomposable.