If the compact complex manifold $X$ has a Hermitian metric $\omega = g(J\cdot, \cdot)$ of negative sectional curvature for the Chern connection, then the first Chern Ricci curvature $$\text{Ric}^{(1)}(\omega) =_{\text{loc}} \sqrt{-1} g^{k \bar{\ell}} R_{i \bar{j} k \bar{\ell}} dz^i \wedge d\bar{z}^j = - \sqrt{-1} \partial \bar{\partial} \log(\omega^n)$$ is negative, and hence, $\eta := - \text{Ric}^{(1)}(\omega)$ defines a closed real $(1,1)$--form that is positive-definite, i.e., a Kähler form. Hence, the manifold must be Kähler. In fact, since the first Chern Ricci curvature is the curvature form of a Hermitian metric on the anti-canonical bundle, negative (Chern) sectional curvature of a Hermitian metric forces the canonical bundle $K_X$ to be a positive line bundle. Kodaira's embedding theorem then implies that $X$ must be projective.
If the compact complex manifold admits a Riemannian (not compatible with the complex structure) metric of negative sectional curvature, then it is not at all clear that such a manifold must be Kähler. It is known that if a compact Kähler manifold is homotopic to a compact Riemannian manifold with negative sectional curvature then it has ample canonical bundle (and is therefore, projective), see this paper.
Some further results: Carlson--Toledo showed that if $N$ is a compact manifold of constant negative curvature and real dimension $4$, then $N$ has no complex structure.