I can give you a partial answer- a reformulation of your question. Perhaps others will then be able to give a counterexample. (I think it is *not* true in general.) Let's start from the beginning.

A map $h : M \to \mathbb C$ is $J$-holomorphic iff $h_* (J Y) = i h_* (Y)$ for every vector field $Y$ on $M$. Equivalently, $dh(JY) = i dh(Y)$, or $(JY) (h) = i Y(h)$, or $(Y
+ i JY)h = 0$, or $(Y - iJY)h = 2Yh$ for any $Y$.

The condition that $X$ is an infinitesmal automorphism of the almost complex structure $J$ is that $[X,JY] = J[X,Y]$ for all $Y$. We want to investigate the question of under what conditions on $J$, for such an $X$, is it true that $(X- i JX)f = 2 Xf$ will be $J$-holomorphic for any $J$-holomorphic $f$.

Ignoring the factor of $2$, we see that
\begin{equation*}
Xf \text{ is $J$-holomorphic} \Leftrightarrow (Y + iJY)(Xf) = 0 \text{ for all $Y$}
\end{equation*}
\begin{equation*}
\Leftrightarrow Y(Xf) = -i(JY)(Xf) \text{ for all $Y$}
\end{equation*}

Recall the definition of the Nijenhuis tensor $N_J(X,Y) = [JX, JY] - J[JX, Y] - J[X,JY] - [X,Y]$. Since $X$ is an infinitesmal automorphism of $J$, the last two terms cancel. Applying both sides of this equation to a $J$-holomorphic map $f$, we get
\begin{equation*}
N_J(X,Y) f = (JX)((JY)f) - (JY)((JX)f) - (J[JX,Y])f = (JX)(iYf) - (JY)(iXf) - i [JX,Y]f
\end{equation*}
\begin{equation*}
= i \left( (JX) (Yf) - (JY)(Xf) - (JX)(Yf) + Y((JX)f) \right)
\end{equation*}
\begin{equation*}
= i \left( - (JY)(Xf) + i Y(Xf) \right)
\end{equation*}
which is exactly the quantity that we need to vanish.

Therefore, $Xf$ will also be $J$-holomorphic iff $N_J(X,Y)f = 0$ for all vector fields $Y$. In particular, this always holds if $J$ is integrable, since then $N_J$ vanishes identically. I don't know an example where $J$ is not integrable but where such an $X$ and $f$ exist. If there is one (which I would bet there is), then in that case $Xf$ would *not* be $J$-holomorphic.