Does the Lie derivative commute with $\partial$?

This is not true as stated.

Suppose $\alpha$ is a $d$-closed $(p, q)$-form, then $\partial\alpha = 0$, so $\mathcal{L}_X\partial\alpha = 0$. On the other hand,

$$\partial\mathcal{L}_X\alpha = \partial(di_X + i_Xd)\alpha = \partial di_X\alpha = \partial(\partial + \bar{\partial})i_X\alpha = \partial\bar{\partial}i_X\alpha.$$

Now let $M = \mathbb{C}$, $\alpha = dz$ and $X = |z|^2\partial_z$. Then we have

$$\partial\mathcal{L}_X\alpha = \partial\bar{\partial}i_X\alpha = \partial\bar{\partial}(i_{|z|^2\partial_z}dz) = \partial\bar{\partial}(dz(|z|^2\partial_z)) = \partial\bar{\partial}|z|^2 = dz\wedge d\bar{z} \neq 0.$$

I don't know if there is a complex analogue of the identity $\mathcal{L}_X d\alpha = d\mathcal{L}_X\alpha$ which would allow one to replace $d$ by $\partial$ or $\bar{\partial}$.

It's true if the complex structure is invariant under the vector field X. In that case: $\mathcal{L}_X \partial a = \mathcal{L}_X (1 - \imath J) da = (1 - \imath J)\mathcal{L}_X da = \partial \mathcal{L}_X a$. (Note: This is written with $a$ being a scalar. For a $(p,q)$ form you just need to pick a suitable projector.)