This should follow from Minsky's paper The classification of Kleinian surface groups, I: Models and bounds on a priori bounds for surface groups, which is used in the proof of the ending lamination conjecture.
The punctured torus case (The classification of punctured-torus groups, Annals) 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 $\mathrm{Area}(A) \approx \gamma''$ by a Gauss-Bonnet argument (if $V$ were a horocusp, then this would be an equality). But
$$
\mathrm{length}(\gamma')\leq \mathrm{length}(\gamma'')\approx \mathrm{Area}(A) \leq \mathrm{Area}(f^{-1}(\Sigma)) = -2\pi \chi(\Sigma).
$$
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 $$
2\pi \mathrm{length}(\gamma')/\mathrm{length}(\mu),
$$
which is very small, and approaches zero as $\mathrm{length}(\gamma)\to 0$.
