NB:  I've had a little time to think about this and can now improve my answer, in particular, removing the real-analytic assumption, which, as I suspected, was not necessary.  Here is the improved answer:

If the metric $g$ is  Kähler, then having the exponential map from a point $p\in M$ be holomorphic makes it flat in a neighborhood of $p$.

Suppose that $\exp_p:T_pM\to M$ is holomorphic near $0_p\in T_pM$ (where we use the natural holomorphic structure on the complex vector space $T_pM$).  Let $z:T_pM\to\mathbb{C}^n$ be a complex linear isometry, so that the hermitian metric on $T_pM$ is just $|z|^2$ in the usual sense.  Let $Z$ be the holomorphic 'radial' vector field on $\mathbb{C}^n$, whose real part is the standard radial vector field on $\mathbb{C}^n$.  

Then
$$
{\exp_p}^*g = g_{i\bar j}(z)\ dz^i\ d\overline{z}^j
$$
for some functions $g_{i\bar j}$ on a neighborhood of $0\in\mathbb{C}^n$.  Since $g$ is Kähler, there is a function $f$ defined on a neighborhood of $0\in\mathbb{C}^n$ such that
$$
g_{i\bar j} = \frac{\partial^2f}{\partial z^i\ \partial\overline{z}^j}.
$$

Now, the condition that $z$ furnish Gauss normal coordinates for ${\exp_p}^*g$ is easily seen to be that  
$$
\mathcal{L}_Z\bigl(\bar\partial f\bigr) = \bar\partial\bigl(|z|^2\bigr).
$$
In particular, $ \bar\partial\bigl(\mathcal{L}_Z(f - |z|^2)\bigr) = 0$, 
so $\mathcal{L}_Z(f - |z|^2) = h$ for some holomorphic function $h$ on a neighborhood of $0$.  This $h$ must vanish at $0$, so it is easy, by adding the appropriate holomorphic function to $f$ (which won't change $g$) to arrange that $h\equiv0$ and, moreover, that $f(0) = 0$.  But this now implies that the real-valued function $f-|z|^2$ vanishes at the origin and also is constant along the radial vector field.  Thus, $f = |z|^2$, and the metric $g$ is flat in these coordinates.