This should follow from Minsky's work on a priori bounds for surface groups, which is used in the proof of the ending lamination conjecture.
The punctured torus case is simpler and more explicit (see Theorem 4.1 and equations 4.4 and 4.5).

Addendum: Once I thought about it for a bit, I think it follows from much more elementary
considerations (in fact, I'm pretty sure someone explained this to me before, but I forgot the argument). Let $\Sigma$ be a surface. Suppose one has a very short geodesic $\gamma\subset M$, where $M\cong \Sigma\times \mathbb{R}$ is a hyperbolic manifold, then Otal's argument proves it is unknotted (this was actually known to Thurston, and generalized to multiple components by Otal). In fact, one may find a pleated surface $f:\Sigma \to M$ so that $\gamma$ is a closed geodesic on the image of this surface. Then the Margulis tube $V$ of $\gamma$ is of very large radius, and therefore its boundary $\partial V$ is very close to being a horosphere (i.e., its principle curvatures are very nearly $=1$) and is isometric to a Euclidean torus. The boundary slope $\gamma'\subset \partial V$ of the surface $\Sigma$ is a Euclidean geodesic of bounded length - this follows from an area estimate of a pleated annulus $A \subset \Sigma$ such that $f(A)$ cobounds $\gamma$ and $\gamma''$, where $\gamma''\sim \gamma'\subset \partial V$, which has $Area(A) \approx \gamma''$ by a Gauss-Bonnet argument (if $V$ were a horocusp, then this would be an equality). But
$$length(\gamma')\leq length(\gamma'')\approx Area(A) \leq Area(f^{-1}(\Sigma)) = -2\pi \chi(\Sigma).$$ The meridian $\mu\subset \partial V$ is a curve intersecting $\gamma'$ once. We may assume that $\gamma',\mu\subset \partial V$ are chosen to be Euclidean geodesics. Then $\partial V \backslash (\gamma'\cup \mu)$ is a Euclidean parallelogram, with one pair of sides of bounded length corresponding to $\gamma'$. Since $V$ has very large radius, $\mu$ must be extremely long. The rotational part corresponds to the fraction of the offset between the two sides of the parallelogram corresponding to $\mu$.
But this implies that the rotational part of $\gamma$ is less than $$2\pi length(\gamma')/length(\mu),$$ which is very small, and approaches zero as $length(\gamma)\to 0$.