Let $M$ be a complex manifold and $\Omega$ a Hermitian differential $(1,1)$-form on $M$. Is it true that for every $x \in M$ there exists an open neighborhood $x \in U \subseteq M$, a local holomorphic frame $v_1,\dots,v_n \in \Gamma(U,T^{1,0}(M))$ and some smooth functions $\Omega_1,\dots,\Omega_n \colon U \to \mathbb{C}$ such that for every $y \in U$ we have that $\Omega_y = \sum_{j=1}^n \Omega_j(y) \cdot v_j^\ast(y) \wedge \overline{v}_j^\ast(y)$, where $\overline{v}_1,\dots,\overline{v}_n \in \Gamma(U,T^{0,1}(M))$ denotes the conjugate frame, and $v_1^\ast,\dots,v_n^\ast \in \mathcal{A}^{1,0}(U)$ are the differential forms dual to $v_1,\dots,v_n$?
This is known to be true under the assumption that $\Omega_x$ is non-degenerate. Indeed, in this case we can choose $U$ so that $\Omega_y$ is non-degenerate for every $y \in U$, and apply Gram-Schmidt paying attention to isotropic vectors, as explained in Proposition 2.66 of Lee's book "Introduction to Riemanniann manifolds". In this case, one can clearly take the functions $\Omega_1,\dots,\Omega_n$ to be constantly equal either to $1$ or to $-1$.
If $\mathrm{rk}(\Omega_x) = r < n$, we can certainly find an open neighborhood $x \in U \subseteq M$ such that $\mathrm{rk}(\Omega_y) \geq r$ for every $y \in U$. If we can find such an open in which $\mathrm{rk}(\Omega_y) = r$ for every $y \in U$, I believe that one can quotient by the totally degenerate subspace, which varies smoothly because the rank is fixed, and conclude by the previous observation.
Can this reasoning be extended to the general case? This seems to have been explained in this post, but no reference for a proof is provided.