It turns out that it is nice to read books. The answer to the weaker version of my question with $O|z|^{n+1}$ term is contained on page 17 of McDuff-Salamon book [MS] (no need of Micallef-White!):
https://people.math.ethz.ch/~salamon/PREPRINTS/jholsm.pdf
Proof. In the proof of Lemma 2.2.3 of [MS] one uses coordinates in $\mathbb C^2$ such that $C_1$ is given by $w=0$ and the almost complex structure $J$ along the line $(z,0)$ is the standard one. Then it is explained that the almost complex map $z\to \mathbb C^2$ corresponding to $C_2$ is given by
$$z\to (p(z)+O(|z^{n+1}|), az^n+O(|z^{n+1}|))$$
where $p(z)$ is a polynomial of order at most $n$, $a\ne 0$. In our case of course $p'(0)\ne 0$. It is now clear that in these coordinates $C_2$ is as need. QED.
Comment. The above proof is elementary and does not use Micallef-White. Similarly to Micallef-White's, statement it can be used to answer the original question with a $C^1$-smooth change of coordinates (instead of $C^{\infty}$). Indeed, after a smooth reparameterization in $z$ and scaling in $w$ the above map for $C_2$ looks as
$$z\to (z, z^n+O(|z^{n+1}|)).$$
Denote the second term by $f(z)$. Then the map
$(z,w)\to (z,w-(f(z)-z^n)/z^n)$ is $C^1$ and it sends the couple $C_1,C_2$ to the couple $(w=0, w=z^n)$.
I wonder still if one can make this last change of coordinates $C^{\infty}$...