Suppose $J :TM \to TM$ is an almost complex structure on $M$, and let $\varphi : TM \to TM$ be a bundle automorphism. Set $J_{\varphi} = \varphi\circ J\circ \varphi^{-1}$, then
\begin{align*}
J_{\varphi}\circ J_{\varphi} &= \varphi\circ J\circ \varphi^{-1}\circ\varphi\circ J\circ \varphi^{-1}\\
&= \varphi\circ J\circ J\circ \varphi^{-1}\\
&= \varphi\circ (-\operatorname{id}_{TM})\circ\varphi^{-1}\\
&= -\varphi\circ\operatorname{id}_{TM}\circ\varphi^{-1}\\
&= -\operatorname{id}_{TM}
\end{align*}
where we have used the fact that $\varphi$ is linear on fibres in the penultimate equality.
Therefore, if $M$ admits an almost complex structure $J$, every bundle automorphism $\varphi$ gives rise to another almost complex structure $J_{\varphi}$, but not every almost complex structure on $M$ arises in this way; for example, $-J$.
Note, the previous paragraph is nothing more than a global version of the following linear algebra statement: if $J$ is a matrix which squares to $-I_n$, then any matrix which is similar to $J$ also squares to $-I_n$, but not every such matrix is similar to $J$; for example, $-J$.
The classic example of an almost complex manifold that is not a complex manifold is the six-sphere $S^6$. Consider $S^6$ in $\mathbb{R}^7 = \mathrm{im}\,\mathbb{O}$ as the set of unit norm imaginary octonions. The almost complex structure on $S^6$ is defined by $J_p v = p \times v$, where $p\in S^6$ and $v\in T_p S^6$ and $\times$ stands for the vector product on $\mathbb{R}^7$.
This almost complex structure cannot be induced by a complex atlas on $S^6$ because the Nijenhuis tensor $N_J$ doesn't vanish (cf. the Nirenberg Newlander theorem).
A result of Borel-Serre states that the only spheres endowed with an almost complex structure are $S^2$ and $S^6$. (The one on $S^2$ is a complex structure.) Up to now, it is not known whether there exists a complex structure on $S^6$.
As a reference I mention the Wikipedia page on almost complex manifolds.
Best Answer
With respect to $J$ we get a splitting of the complexified differential forms; namely,
$$\Omega^k(X)\otimes\mathbb{C} = \bigoplus_{p+q=k}\Omega^{p,q}(X).$$
With respect to this splitting, we have the operator $\overline{\partial} : \Omega^{p,q}(X) \to \Omega^{p,q+1}(X)$ given by $\pi^{p,q+1}\circ d$ where $d$ is the exterior derivative and $\pi^{p,q+1}$ is the projection $\Omega^{k+1}(X)\otimes\mathbb{C} \to \Omega^{p,q+1}(X)$.
Integrability of $J$ is equivalent to the condition $\overline{\partial}^2 = 0$, but this is automatic on a real surface because we have
\begin{align*} \Omega^0(X)\otimes\mathbb{C} &= \Omega^{0,0}(X)\\ \Omega^1(X)\otimes\mathbb{C} &= \Omega^{1,0}(X) \oplus \Omega^{0,1}(X)\\ \Omega^2(X)\otimes\mathbb{C} &= \Omega^{1,1}(X) \end{align*}
and $\Omega^{p,q}(X) = 0$ for all other $p$ and $q$. Now note that
$$\Omega^{0,0}(X) \xrightarrow{\overline{\partial}} \Omega^{0,1}(X) \xrightarrow{\overline{\partial}} \Omega^{0,2}(X) = 0.$$
So $\overline{\partial}^2 = 0$ on $\Omega^{0,0}(X)$. Likewise, $\overline{\partial}^2 = 0$ on $\Omega^{1,0}(X)$, $\Omega^{0,1}(X)$, and $\Omega^{1,1}(X)$.
Added Later: Altenatively, we can show that for any almost complex structure $J$ on a smooth two-dimensional manifold, $N_J = 0$ where $N_J$ is the Nijenhuis tensor field of $J$. By the Newlander-Nirenberg Theorem, the vanishing of $N_J$ is equivalent to integrability.
Let $M$ be a smooth manifold with almost complex structure $J$. Recall that the Nijenhuis tensor field of $J$ is given by
$$N_J(X, Y) = [X, Y] + J[JX, Y] + J[X, JY] - [JX, JY].$$
As $N_J$ is tensorial, if $\{V_1, \dots, V_n\}$ is an ordered basis of local vector fields defined in a neighbourhood of $p$, $N_J$ vanishes at $p$ if and only if $N_J(V_i, V_j) = 0$ for all $1 \leq i, j \leq n$. Furthermore, $N_J$ is skew-symmetric, so it is enough to show that $N_J(V_i, V_j) = 0$ for all $1 \leq i < j \leq n$.
Now consider the case where $M$ is a smooth two-dimensional manifold. Fix $p \in M$ and let $V$ be a nowhere zero local vector field defined in a neighbourhood of $p$, then $\{V, JV\}$ is an ordered basis of local vector fields; to see this, note that if $JV \in \operatorname{span}\{V\}$ then $JV = kV$ so $-V = J^2V = k^2V$ which is impossible. As
\begin{align*} N_J(V, JV) &= [V, JV] + J[JV, JV] + J[V, J^2V] - [JV, J^2V]\\ &= [V, JV] + J[V, -V] - [JV, -V]\\ &= [V, JV] - J[V, V] + [JV, V]\\ &= [V, JV] - [V, JV]\\ &= 0, \end{align*}
we see that $N_J = 0$ at $p$. As $p$ is arbitrary, the Nijenhuis tensor field of $J$ vanishes, so $J$ is integrable.