A classic example that shows that WLNT doesn't always hold in the holomorphic category is an elliptically fibered $K3$ surface.

A K3 surface $S$ is a compact complex symplectic manifold of complex dimension $2$. Any smooth curve $C\subset S$ is a Lagrangian submanifold. If the Darboux Theorem were true in the sense that a neighborhood of the curve were always symplectically biholomorphic with a neighborhood of the zero section of the cotangent bundle, then, for a nonsingular elliptic curve $C\subset S$, the neighborhood would be a product and so the $1$-parameter family of nearby deformations of the curve $C$ in $S$ would all be isomorphic to it, i.e., they would all have the same $j$-invariant. However, this is known not to be the case: When you have an elliptically fibered $K3$, the $j$-invariant of the elliptic fibers is not constant.

If one doesn't need the ambient symplectic manifold to be compact, there is an easier example: Let $z$ be the standard complex coordinate on $\mathbb{C}$ and let $w$ be the standard complex coordinate on the upper half-plane $\mathbb{U}^+\subset\mathbb{C}$. Let $\mathbb{Z}^2$ act (freely) on $X = \mathbb{C}\times\mathbb{U}^+$ by
$$
(m,n)\cdot(z,w) = (z + m + nw,\,w).
$$
Let $M$ be the quotient of $\mathbb{C}\times\mathbb{U}^+$ by this $\mathbb{Z}^2$-action. This action preserves the symplectic form $\mathrm{d}z\wedge\mathrm{d}w$, so that $M$ is a symplectic surface, as well as the projection $w:M\to \mathbb{U}^+$, whose fiber in $M$ over $a\in \mathbb{U}^+$ is a torus $w^{-1}(a) \simeq \mathbb{C}/(\mathbb{Z}{+}\mathbb{Z}a)$. These tori are Lagrangian curves in $M$, but no neighborhood of such a torus is biholomorphic to the product of the torus and a disk, precisely because the tori near a given $w$-fiber (which are all $w$-fibers since $w$ would have to be constant on any compact, connected curve in $M$) are not biholomorphic to the given $w$-fiber.

Maybe the result you want is to know that any symplectic form on a neighborhood of the zero section of the cotangent bundle that has the zero section as a Lagrangian submanifold is actually symplectomorphic to the canonical symplectic form on some neighborhood of the zero section of the cotangent bundle. This statement is true, and the standard homotopy argument proves it. (You don't need to use local coordinates for the proof, just the radial vector field on the cotangent bundle to construct the homotopy.)

**Added Remark:** However, this result is particular for the zero section of the cotangent bundle of $L$. It is *not* in general true that, for any two holomorphic symplectic structures $\omega_1$ and $\omega_2$ on a complex manifold $M$ that have a common holomorphic Lagrangian submanifold $L\subset M$, there exist open $L$-neighborhoods $U_1$ and $U_2$ in $M$ such that there exists a biholomorphic mapping $\phi:U_1\to U_2$ fixing $L$ and satisfying $\phi^*\omega_2 = \omega_1$.

For example, taking the above torus fibration $w: M\to \mathbb{U}^+$, for different nonvanishing holomorphic functions $f_1,f_2:\mathbb{U}^+\to\mathbb{C}$, the symplectic forms $\omega_i = f_i(w)\,\mathrm{d}z\wedge\mathrm{d}w$ for $i=1,2$ will not generally be locally holomorphically symplectomorphic near any $w$-fiber.