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.

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 $$ \partial f \wedge (\overline{z_j}\,\mathrm{d}z_j) = \overline\partial f \wedge (z_j\,\mathrm{d}\overline{z_j}) = 0.\tag1 $$ In particular, these equations will hold since they are the same as $$ \partial f \wedge \partial (|z|^2) =\overline \partial f \wedge \overline\partial (|z|^2) = 0,\tag2 $$ which holds, by hypothesis.

Addendum: In fact, a stronger result holds: If $n\ge 2$ and $f$ is any smooth (real-valued) function on $\mathbb{C}^n$, then the mapping $\phi:\mathbb{C}^n\to\mathbb{C}^n$ defined by $\phi(z) = f(z)z$ pulls back the form $\omega$ to be of type $(1,1)$ if and only if $f$ is constant on the level sets of $|z|$. The proof is a little more subtle in the $n=2$ case, but, basically what you have to observe is that (1) implies that there is a (complex) function $g$ such that $$\mathrm{d}f = g\,(\overline{z_j}\,\mathrm{d}z_j) +\overline{g}\, \overline{z_j}\,\mathrm{d}z_j\,.\tag3 $$ Taking the exterior derivative of (3) yields a relation $$ 0 = \mathrm{d}g\wedge(\overline{z_j}\,\mathrm{d}z_j) +\mathrm{d}\overline{g}\wedge\overline{z_j}\,\mathrm{d}z_j +(\overline{g}-g)\,\omega,\tag 4 $$ which, when $n>2$, implies by elementary exterior algebra that $g = \overline{g}$, so that $\mathrm{d}f = g\mathrm{d}(|z|^2)$, implying that $f$ is constant on the level sets of $|z|$. When $n=2$, one has to examine (4) more carefully, but it still implies that $g = \overline{g}$.