Let $\varphi: (S,J_S) \to (M,J_M)$ be a map from a complex manifold equipped with a integrable complex strucure $J_S$ to a smooth manifold equipped with a almost complex structure $J_M$.
Using $J_M$, the tangent bundle $TM$ of $M$ decomposes to $TM\otimes \mathbb{C} = TM^{1,0} \bigoplus TM^{0,1}$. Then the differential $d\varphi: TS\otimes \mathbb{C} \to TM\otimes \mathbb{C}$ also decomposes to $d \varphi = \partial \varphi \oplus\bar{\partial} \varphi$. If we assume $\bar{\partial} \varphi = 0$, then can we say that $J_M \circ d\varphi = d\varphi \circ J_S $?
In case of $v \in TS^{1,0}$, I can prove it. In fact, $J_Sv=\sqrt{-1}v$ and hence $d \varphi \circ J_Sv=\sqrt{-1} d \varphi (v)$. On the other hand, because $d \varphi (v) \in TM^{1,0}$ we can obtain $J_M \circ d \varphi (v)= \sqrt{-1} d \varphi (v)$. Thus, in this case, $J_M \circ d\varphi = d\varphi \circ J_S $.
In case of $v \in TS^{0,1}$, I cannot prove it.
Do I misunderstand anything?
Note that if $v \in TS^{0,1}$, then $\bar v \in TS^{1,0}$. Since the action of $d\varphi$, $J_S$ on $TS \otimes \mathbb C$ are given by $$d\varphi (X \otimes z) = d\varphi (X) \otimes z,\text{ and }J_S (X\otimes z) = J_S(X) \otimes z,$$ and similar for the action $J_M$ on $TM\otimes \mathbb C$, we have (using your work when $\bar v \in TS^{1,0}$)
\begin{align} d\varphi (J_S (v)) &= \overline{d\varphi (J_S \bar v)}\\ &= \overline {J_M (d\varphi (\bar v)) }\\ &= J_M (d\varphi (v)). \end{align}