This should follow from [Minsky's work][1] on a priori bounds for surface groups, which is used in the proof of the ending lamination conjecture. 
The [punctured torus case][2] 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 $length(\gamma)\to 0$. 

  

  [1]: http://front.math.ucdavis.edu/0302.5208
  [2]: http://www.ams.org/mathscinet-getitem?mr=1689341