If the Hermitian metric is not Kähler, the Laplacian won't respect bi-degree. In fact, a more general result is true: If an almost Hermitian metric is not Kähler, then its Laplacian will not respect bi-degree.
Here's a simple explicit (i.e., 'concrete') example: Let $X = \mathbb{C}^2$ with standard holomorphic coordinates $z_1 = x_1 + i\,x_2$ and $z_2=x_3 + i\,x_4$. Now let $f$ be a smooth function on $X$ and consider the positive definite $(1,1)$-form
$$
\omega = \mathrm{e}^{2f}\mathrm{d}x_1\wedge \mathrm{d}x_2 + \mathrm{d}x_3\wedge \mathrm{d}x_4
$$
with associated Hermitian metric $g = \mathrm{e}^{2f}({\mathrm{d}x_1}^2 + {\mathrm{d}x_2}^2) + {\mathrm{d}x_3}^2 + {\mathrm{d}x_4}^2$.
It's a straightforward calculation to show that $\Delta\omega$ has type $(1,1)$ if and only if
$$
\frac{\partial^2f}{\partial x_2\partial x_3}
+\frac{\partial^2f}{\partial x_1\partial x_4}
= \frac{\partial^2f}{\partial x_1\partial x_3}
-\frac{\partial^2f}{\partial x_2\partial x_4} = 0.
$$
(These are the conditions that the (2,0) and (0,2) pieces of $\Delta\omega$ vanish.) The point is that, because $\omega = \ast\omega$, the formula for the Laplacian simplifies to
$$
\Delta\omega = -d{*}d{*}\omega-*d{*}d\omega = -(1+*)(d{*}d\omega),
$$
so $\Delta\omega$ is minus twice the projection of $d{*}d\omega$ into its self-dual part, which, since it is self-dual, is the sum of a multiple of $\omega$ itself with the real part of a $(2,0)$-form. The above two equations represent the real and imaginary parts of the coefficient of the $(2,0)$ part.
To answer the OP's final question, I'll state the following result, which is stronger than needed:
Proposition: Let $(M,J,g)$ be an almost Hermitian manifold and suppose that its $g$-Laplacian $\Delta$ carries $(0,1)$-forms to $(0,1)$-forms. Then $J$ is $g$-parallel, i.e., $(M,J,g)$ is Kähler.
Proof: (sketch) Let $\alpha\in\Omega^1(M)$ be any $1$-form on $M$. Define $J^*\alpha$ by the property $J^*(\alpha)(v) = \alpha(Jv)$. Since, by hypothesis, $\Delta$ carries $(0,1)$-forms to $(0,1)$-forms, it follows that $\Delta$ carries $(1,0)$-forms to $(1,0)$-forms, and in particular, it follows that $\Delta(J^*\alpha) = J^*(\Delta\alpha)$.
Now, it's a straightforward fact that if $L:TM\to TM$ is any smooth vector bundle map (it doesn't have to be a complex structure or compatible with $g$), then $\Delta(L^*\alpha) = L^*(\Delta\alpha)$ holds for all $1$-forms $\alpha$ if and only if $L$ is parallel with respect to the metric $g$, i.e., its covariant derivative with respect to $g$ (when $L$ is thought of as a section of $TM\otimes T^*M$), vanishes. [The point is that the difference $\Delta(L^*\alpha) - L^*(\Delta\alpha)$ is first order in $\alpha$, and it's enough to check what the condition is on the covariant derivatives of $L$ at one point $p$ when one considers the 1-forms $\alpha$ that vanish at $p$. Then it's easy.]
Once one knows this, then the desired result immediately follows, since an almost Hermitian structure $(M,J,g)$ is Kähler if and only if $J$ is parallel with respect to $g$. $\square$
Note: I had forgot about this argument when I answered this question last December. In fact, I showed this argument to Jeremy Daniel and Xiaonan Ma back in 2013, who incorporated it into their arXiv paper Characteristic Laplacian in sub-Riemannian geometry as Remark 2.3 and Lemma 2.4
