Yes, it's true, and it follows immediately from a simple calculation:  Since $f$ is assumed to be a function of $|z|$ such that $\phi(z) = f(|z|)z$ is a smooth diffeomorphism, it follows that $\mathrm{d}f$ is a multiple of $\mathrm{d}\bigl(|z|^2\bigr)$, so $\partial f$ is a multiple of $\partial\bigl(|z|^2\bigr)$.  Now
$$
\phi^*(\omega) = \tfrac{i}{2} \,\phi^*\left(\mathrm{d}z_j\wedge\mathrm{d}\overline{z_j}\right)
= \tfrac{i}{2} \mathrm{d}(fz_j)\wedge\mathrm{d}(f\overline{z_j})
=  \tfrac{i}{2}\left(f^2 \mathrm{d}z_j\wedge\mathrm{d}\overline{z_j} + f\,\mathrm{d}f\wedge(z_j\,\mathrm{d}\overline{z_j}-\overline{z_j}\,\mathrm{d}z_j) \right).
$$
The (2,0) and $(0,2)$ parts of the right hand side vanish if and only if 
$$
\partial f \wedge (\overline{z_j}\,\mathrm{d}z_j) 
= \overline\partial f \wedge (z_j\,\mathrm{d}\overline{z_j}) = 0.
$$
In particular, these equations will hold since they are the same as 
$$
\partial f \wedge \partial (|z|^2) =\overline \partial f \wedge \overline\partial (|z|^2) = 0,
$$
which holds, by hypothesis.