The answer to your first question is yes -- regardless of whether the metric $g$ is compatible with $J$, the section $H \in \Gamma(T^*X^{\otimes 2} \otimes \mathbb{C})$ defined by $H(\lambda v, \mu w):=\tfrac{1}{2}\lambda\overline{\mu}g(v,w)$ is a Hermitian metric. This is just because it is a smoothly varying Hermitian inner product on each fiber of $TX \otimes \mathbb{C}$. That having been said, $H$ will not necessarily play nicely with the holomorphic and antiholomorphic bundles, commonly denoted by $T^{1,0}X$ and $T^{0,1}X$. In fact, these two subbundles are $H$-orthogonal to one another if and only if $g$ is $J$-compatible.
In the case that $g$ is not $J$-compatible, we will also no longer have the nice decomposition $H= g+i\omega$, which is the standard shorthand for the situation (indeed, there is no reason for the form define by
$$(v+iJv,w+iJw) \mapsto g(v,w) + ig(Jv,w)$$
to be Hermitian if $g$ is not assumed to be compatible with $J$).
We will, however, still have an analogous decomposition $H= \tilde{g} + i\tilde{\omega}$, where
$$\tilde{g}(v,w):=\tfrac{1}{2}\big(g(v,w)+g(Jv,Jw)\big)$$
and
$$\tilde{\omega}(v,w):=\tfrac{1}{2}\big(g(Jv,w)-g(v,Jw)\big).$$
As before, we have that $g=\tilde{g}$ and $\omega=\tilde{\omega}$ iff $g$ is compatible with $J$.
Please let me know if I can help clarify any of this. Differential geometry is rife with abuses of notation. Of course, as soon as one understands what is meant by the abusive notation, one happily adopts it. I am no exception =).
If $\nabla$ is any connection and $f$ a function, its Hessian with respect to $\nabla$ is $\mathrm{Hess}^{\nabla}f = \nabla \mathrm{d}f$, and one can see, after a messy calculation, that:
$$
\mathrm{Hess}^{\nabla}f(X,Y) - \mathrm{Hess}^{\nabla}f(Y,X) = \pm\mathrm{d}f\left([X,Y] - (\nabla_XY - \nabla_YX) \right)
$$
(where the $\pm$ sign is here because I don't remember the exact sign, but the computations are not that hard, just messy.) Hence, Hessians are symmetric if and only if the connection is torsion-free. This is the main motivation to consider torsion-free connections: in the euclidean space, Hessians are symmetric!
Moreover, the fundamental theorem of Riemannian geometry tells us that on a Riemannian manifold, there is a unique connexion that is torsion-free and lets the metric invariant, that is:
$$
\forall X,Y,Z, \left(\nabla_Zg\right)(X,Y) = Z\cdot g\left(X,Y \right) - g\left(\nabla_ZX,Y\right) - g\left(X,\nabla_ZY\right) = 0.
$$
(compare with the euclidean case, where $\langle X,Y\rangle ' = \langle X',Y\rangle + \langle X, Y' \rangle$.)
This theorem thus says that given any Riemannian metric $g$, there is a connection that is better than others: Hessians are symmetric and the metric is invariant under the action. We call it the Levi-Civita connexion.
If a connection is chosen, a geodesic is a parametrized curve satisfying the equation of geodesics : $\nabla_{\gamma'}\gamma' = 0$. Thus a curve $\gamma$ is a geodesic with respect to the connection, and can be a geodesic for some connection $\nabla^1$ but not for another connecion $\nabla^2$. Therefore, your question does not really have sense: we do not say that a connexion gives the least energy of a geodesic. I think you got confused, believing that being a geodesic is an intrinsic notion, but it really depends on the connection you consider.
Now, suppose $(M,g)$ is a Riemannian manifold endowed with its Levi-Civita connexion. Then if $\gamma : [a,b] \to M$ is a curve, we define its energy to be:
$$
E(\gamma) = \frac{1}{2}\int_a^b \|\gamma'\|^2
$$
and one can show that, in the space of all curves $\{\gamma : [a,b] \to M\}$ with same end points, a curve $\gamma$ is a point where the energy functional is extremal if and only if $\nabla_{\gamma'}\gamma'=0$, that is if and only if $\gamma$ is a solution of the equation of geodesics. Hence, a minimizer of the energy functional is a geodesic.
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