You may already know this, and this is not the most general statement, probably, but it does work for all almost complex structures (integrable or not):

First, some notation: Using the almost complex structure $J$ on $X$, split the exterior derivative into graded pieces $d^{s,t}:\Omega^{p,q}(X)\to\Omega^{p+s,q+t}(X)$, where $(s,t)\in\{(2,-1),(1,0),(0,1),(-1,2)\}$. Of course, $d^{2,-1}$ and $d^{-1,2}$ are tensorial (i.e., zeroth order linear differential operators) and vanish when $J$ is integrable. Maintain the standard abbreviations $d^{1,0}=\partial$ and $d^{0,1}=\overline\partial$. [Note that, because pseudo-holomorphic mappings preserve bi-degree (and, of course, 'commute' with the exterior derivative), these operators commute with pullback via pseudo-holomorphic maps.]

Given a (say, smooth) function $f:X\to\mathbb{R}$ define the *Levi form* of $f$ to be
$$
{\mathsf L}(f) = i\,\partial\overline{\partial}\,f \in \Omega^{1,1}(X).
$$
Then ${\mathsf L}(f)$ is a real-valued $(1,1)$-form on $X$. One says that $f$ is *plurisubharmonic* on $X$ if ${\mathsf L}(f)\ge0$ (in the sense that it is nonnegative on all (oriented) $1$-dimensional complex subspaces of $TX$).

**Proposition:** Let $f:X\to\mathbb{R}$ be plurisubharmonic. If $u:\Delta\to X$ is $J$-holomorphic and $f\bigl(u(z)\bigr)\le f\bigl(u(0)\bigr)$ for all $z\in\Delta$, then $f\bigl(u(z)\bigr) =f\bigl(u(0)\bigr)$ for all $z\in \Delta$.

The relation with the OP's question is this: If $X\subset\mathbb{C}^n$ and $c$ is a regular value of $f:X\to\mathbb{R}$, where $f$ is plurisubharmonic, then $H_c=f^{-1}(c)$ is weakly pseudo-convex.

It's not clear to me that a smooth weakly pseudo-convex hypersurface in $X$ is locally a level set of a plurisubharmonic function $f$, but it seems plausible. (See the comment below, which asserts that this *is* true in the integrable case, by a result of Diederich-Fornæss.)

The proof of the above proposition follows from the (usual) maximum principle on $\Delta$ and the formula
$$
\frac{\partial^2(f{\circ}u)}{\partial z\,\partial \overline z}
= \frac1i \,u^*\bigl({\mathsf L}(f)\bigr)\left(\frac{\partial}{\partial z},\frac{\partial}{\partial \overline{z}}\right) \ge0,
$$
which is a consequence of the fact that the Levi form operator commutes with pullback under pseudo-holomorphic maps. (Note, in particular, that the Nijnhuis tensor does not affect the formula.)