It is known (ScwhartzPick) that holomorphic functions are distance nonincreasing in the hyperbolic metric.Does it imply that they preserve convexity in the hyperbolic metric? How about the converse?
No. Take a convex set and use an exponential function $e^{2\pi i z/c}$, where $c$ is the difference between two points in the set. The image of the set is no longer simply connected, and thus cannot be convex in a Riemannian manifold with no closed geodesics like the hyperbolic disc (which I assume is what you are talking about here). You may need to scale the exponential function down by a constant to fit in the disc, this is obviously immaterial
@Will gave an excellent answer, but also, a mapping preserving convexity must map (hyperbolic) straight lines to (hyperbolic) straight lines (since the images of both the halfspaces need to be convex), and the only analytic functions which do that are Mobius transformations, so the answer is NO almost always.

$\begingroup$ If the function is bijective, one does not even need analyticity, according to the reference: Demirel, Oğuzhan; Soytürk Seyrantepe, Emine A characterization of Möbius transformations by use of hyperbolic regular polygons. J. Math. Anal. Appl. 374 (2011), no. 2, 566–572 They prove that a continuous bijection of the unit disk is Moebius if and only if it preserves hyperbolic regular polygons. $\endgroup$ – Margaret Friedland Jun 13 '12 at 17:35