TL;DR: Given a (possibly non-symmetric) non-degenerate covariant (0,2) tensor field $$\mathbb{g}~\in~\Gamma(T^{\ast}M\otimes T^{\ast}M),\tag{1}$$
then a compatible torsionfree tangent-bundle connection $\nabla$ might not exists or be unique. See e.g. the Theorem below.
Theorem. Given a non-degenerate 2-form $$\frac{1}{2}\omega_{ij}\mathrm{d}x^i\wedge\mathrm{d}x^j~=~\omega~\in~\Omega^2(M)~\equiv~\Gamma(\bigwedge\!{}^2T^{\ast}M)\tag{2}$$
on a paracompact manifold $M$. Define
$$\frac{1}{3!}\Omega_{ijk}\mathrm{d}x^i\wedge\mathrm{d}x^j\wedge\mathrm{d}x^k~=~\Omega~:=~\mathrm{d}\omega~\in~\Omega^3(M).\tag{3}$$ Then a compatible torsionfree tangent-bundle connection $\nabla$ exists iff $\Omega=0$. In the affirmative case, there are infinitely many such connections.
Remark. Such connections are called Fedosov connections, cf. e.g. this Phys.SE post.
Sketched proof of Theorem. One may show (via partition of unity) that there exists a torsionfree connection $\nabla^{(0)}$ on a paracompact manifold $M$. We search for a compatible torsionfree connection $\nabla$. The difference in Christoffel symbols $$\Delta\Gamma^k_{ij}~:=~\Gamma^k_{ij}-\Gamma^{(0)k}_{ij}~=~(i\leftrightarrow j)\tag{4}$$
is a $(1,2)$ tensor. If we define a $(0,3)$-tensor
$$\Delta\Gamma_{k,ij}~:=~\omega_{k\ell}\Delta\Gamma^{\ell}_{ij}~=~(i\leftrightarrow j),\tag{5}$$
the compatibility condition
$$\nabla_k\omega_{ij}~\stackrel{?}{=}~0\tag{6}$$
then turn into the tensor equation
$$\begin{align} -\nabla^{(0)}_k\omega_{ij}~\stackrel{?}{=}~&(\nabla_k-\nabla^{(0)}_k)\omega_{ij}\cr
~=~&-\Delta\Gamma^{\ell}_{ki}\omega_{\ell j}-\omega_{i \ell}\Delta\Gamma^{\ell}_{kj}\cr
~=~&\Delta\Gamma_{j,ki}-\Delta\Gamma_{i,kj}.\end{align}\tag{7}$$
If we define a $(0,3)$-tensor
$$ S_{k,ij}~:=~\Delta\Gamma_{k,ij}-\frac{1}{3}\left(\nabla^{(0)}_i\omega_{kj}+\nabla^{(0)}_j\omega_{ki} \right)~=~(i\leftrightarrow j),\tag{8}$$
then eq. (7) becomes
$$\begin{align} \frac{1}{3}\Omega_{ijk}
~=~&\frac{1}{3}\sum_{ijk~{\rm cycl.}}\partial_i\omega_{jk}\cr
~=~&\frac{1}{3}\sum_{ijk~{\rm cycl.}}\nabla^{(0)}_i\omega_{jk}\cr
\stackrel{?}{=}~&S_{i,kj}-S_{j,ki}.\end{align}\tag{9}$$ If $g_{ij}$ is a positive definite diagonal metric (in one coordinate system), then eq. (9) implies
$$ 0~\leq~\Omega_{ijk}(g^{-1})^{i\ell}(g^{-1})^{jm}(g^{-1})^{kn}\Omega_{\ell mn}~\stackrel{?}{=}~0,\tag{10}$$
which in turn implies that $$\Omega~=~0.\tag{11}$$ Conversely if $\Omega=0$, then eq. (9) becomes
$$S_{i,kj}~\stackrel{?}{=}~S_{j,ki},\tag{12} $$
i.e., that
$$S~\stackrel{?}{\in}~{\rm Sym}^3T^{\ast}M\tag{13}$$
is a totally symmetric tensor. But there are infinitely many such totally symmetric tensors $S$, cf. my Phys.SE answer here. So the sought-for connection $\nabla$ is far from unique.
$\Box$
References:
- B.V. Fedosov, Deformation Quantization and Index Theory, Mathematical Topics, Vol. 9, Akademie Verlag, Berlin, 1996; Proposition 2.5.2.