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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.