Update: I have had a little time this weekend to think more deeply about the construction I outlined below and I have concluded that, unfortunately, the construction that I thought would work to produce a cocompact free action of a discrete subgroup of $G$ on the quotient $G/K$ does not seem to work. (At least, it would require a different method than what I proposed, starting with a cocompact discrete subgroup of $\mathrm{SL}(2,\mathbb{R})$.) Thus, constructing a compact example by this method will not work easily. I'm not removing the original answer entirely since some people have upvoted it (of course, they might want to reconsider their upvotes, given that this method didn't succeed).
Meanwhile, I have decided that there is a better method to try, but am still working on the details, which involve choosing a uniform (i.e., cocompact) discrete lattice in $\mathrm{SL}(3,\mathbb{R})$ with certain properties that I still hope will be possible.
The reason $\mathrm{SL}(3,\mathbb{R})$ enters the picture is this: If $X$ is a complex surface endowed with a foliation $\mathcal{F}$ by holomorphic curves, each point $x\in X$ has an open neighborhood $U$ on which there exists a nonvanishing $(1,0)$-form $\omega$ (unique up to a nonzero multiple) with the property that $\omega=0$ defines the complex tangent lines to the foliation $\mathcal{F}$ in $U$. The foliation is holomorphic in $U$ if and only if $\omega\wedge d\omega\equiv0$ (this is independent of the choice of $\omega$). Say that $\mathcal{F}$ is nowhere holomorphic if $\omega\wedge d\omega$ is nonvanishing for all such $x$ and $U$.
Now, by standard techniques, one can show that if $\mathcal{F}$ is nowhere holomorphic, then the sheaf of vector fields on $X$ whose flows preserve both the complex structure and $\mathcal{F}$ has finite dimensional stalks; in fact, the dimension can never be more than $8$, and it equals $8$ everywhere if and only if $(X,\mathcal{F})$ is locally equivalent to a standard model $(X_0,\mathcal{F}_0)$ defined as follows:
Let $\mathrm{SL}(3,\mathbb{R})\subset\mathrm{SL}(3,\mathbb{C})$ act in the usual way on $\mathbb{CP}^2$. Of course, this action preserves the real slice $\mathbb{RP}^2\subset\mathbb{CP}^2$ and is transitive on the (open) complement of this real slice. In fact, this action commutes with the natural conjugation $[v]\mapsto[\bar v]$, and for any $[v]\in \mathbb{CP}^2\setminus\mathbb{RP}^2$, there is a unique complex line $[v\bar v]$ passing through both $[v]$ and $[\bar v]$, and $X_0=\mathbb{CP}^2\setminus\mathbb{RP}^2$ is foliated by such `real' lines (the leaves in $X_0$ are Poincaré disks). It is not difficult to see that this foliation $\mathcal{F_0}$ is nowhere holomorphic and that $\mathrm{SL}(3,\mathbb{R})$ acts transitively and holomorphically on $X_0$ preserving $\mathcal{F_0}$. Moreover, any local biholomorphism of a connected open subset $U\subset X_0$ into $X_0$ that preserves $\mathcal{F}_0$ is the restriction to $U$ of the action of an element of $\mathrm{SL}(3,\mathbb{R})$.
What seems possible (but, again, I need to check details and I'm not an expert on discrete subgroups of higher rank Lie groups of split type) is that there might exist a lattice $\Gamma\subset \mathrm{SL}(3,\mathbb{R})$ such that $\Gamma$ preserves a connected open subset $U\subset X_0$ on which it acts freely and properly discontinuously and such that the quotient $M=\Gamma\backslash U$ is compact. If so, that will provide an example of what we are looking for.
Original Attempt (modified): A locally homogeneous (noncompact) example can be constructed as follows: First, let $G\subset\mathrm{SL}(3,\mathbb{C})$ be the $5$-dimensional, connected Lie subgroup whose Lie algebra is $$ {\frak{g}} = \left\{\ \begin{pmatrix}ir & z & w\\ \bar z & -ir & \bar{w}\\0&0&0\end{pmatrix}\ \ : \ r\in\mathbb{R},\ z,w\in\mathbb{C}\ \right\}, $$ let $H\subset G$ be the connected subgroup whose Lie algebra is as above with $w=0$, while $K\subset H$ is the connected Lie subgroup whose Lie algebra is as above with $z=w=0$. Then $H\simeq\mathrm{SL}(2,\mathbb{R})$ while $K\simeq S^1 = \mathrm{SO}(2)$.
Now consider the (non compact) $G$-homogeneous space $M = G/K$. I claim that there is a $G$-invariant complex structure on $M$ and a $G$-invariant nonholomorphic foliation of $M$ by complex curves. To see this, consider the canonical left invariant $1$-form on $G$, written in the form $$ \gamma = g^{-1}\ dg = \begin{pmatrix}i\rho & \zeta & \omega\\ \bar\zeta & -i\rho & \bar{\omega}\\0&0&0\end{pmatrix}, $$ where $\rho$ is $\mathbb{R}$-valued and $\zeta$ and $\omega$ are $\mathbb{C}$-valued. The structure equation $d\gamma = -\gamma\wedge\gamma$ gives $$ d\begin{pmatrix}\zeta \\\omega\end{pmatrix} = -\begin{pmatrix}2i\rho&0 \\-\bar\omega& i\rho\end{pmatrix}\wedge \begin{pmatrix}\zeta \\\omega\end{pmatrix} $$ as well as $d\rho = i\ \zeta\wedge\bar\zeta$.
From these equations it follows immediately that there is unique complex structure $J$ on $M=G/K$ for which the $(1,0)$-forms on $M$ pullback to $G$ to be linear combinations of $\zeta$ and $\omega$, that this complex structure is $G$-invariant, and that there is a unique foliation $\mathcal{F}$ of $M$ by $J$-holomorphic curves such that preimages in $G$ of the leaves of $\mathcal{F}$ are the leaves of the codimension $2$ foliation defined by $\omega=\bar\omega=0$. Moreover, because $\omega\wedge d\omega\not=0$, the foliation $\mathcal{F}$ is not holomorphic.
Now, as I wrote, $M$ is not compact, so it's not the example that you are looking for. However, if we now use the fact that $H\simeq\mathrm{SL}(2,\mathbb{R})$, we can find a discrete subgroup $\Gamma\subset H=\mathrm{SL}(2,\mathbb{R})$ such that $\Gamma$ acts properly and discontinuously on $H/K = \Delta$ (i.e., the Poincaré disk) and such that the quotient $C = \Gamma\backslash (H/K)$ is a compact Riemann surface.
[Here is the step that does not work to produce a desired $\hat\Gamma$:] Choosing $\Gamma$ to be an appropriate congruence subgroup, for example, we can even assume that $\Gamma$ preserves a lattice in $\mathbb{R}^2\simeq\mathbb{C}$. In other words, there exists a lattice $\Lambda\subset\mathbb{C}$ such that the abelian discrete subgroup of $G$ defined as $$ \hat\Lambda = \left\{\ \begin{pmatrix}1 & 0 & w\\ 0 & 1 & \bar{w}\\0&0&1\end{pmatrix}\ \ : w\in\Lambda\ \right\}, $$ is stable under conjugation by $\Gamma\subset H$.
Finally, let $\hat\Gamma\subset G$ be the subgroup generated by $\Gamma\subset H\subset G$ and $\hat\Lambda$. Then I believe (but haven't had time to check the details) that $\hat\Gamma$ acts properly and discontinuously on $M = G/K$ with compact quotient. Since $\hat\Gamma$, being a subgroup of $G$, preserves the complex structure on $M$ and the foliation $\mathcal{F}$, it follows that these pass to the quotient $X = \hat\Gamma\backslash(G/K)$ and give an example of the desired kind.
