**Question:** Is the length of homotopically non-trivial loops in $M(f)$ bounded below in terms of the genus $g$ of the fiber?

**Answer:** No. Here is one family of examples. Suppose that $S$ is the closed oriented surface of genus two. Suppose that $\alpha$ is an essential curve in $S$. Let $T_\alpha$ be the left Dehn twist about $\alpha$. Suppose that $g \colon S \to S$ is a pseudo-Anosov map. Define $f_n = g T_\alpha^n$, for any integer $n$. Fathi proves that $f_n$ is again pseudo-Anosov (for all but at most seven consecutive values of $n$.) The Minsky "machine" (namely the proof of the ending lamination conjecture, due to Minsky and Brock-Canary-Minsky) shows that the injectivity radius of $M(f_n)$ tends to zero as $n$ tends to $\pm \infty$.

Note that, in the above examples, the "shrinking curve" in $M(f_n)$ does not "go around" the bundle: that is, its image in $\mathbb{Z} \cong \pi_1(S^1)$ is trivial. Your question has a positive answer if we avoid such curves.

**Sketch proof**: (without actual estimates, of the contrapositive.) Suppose that $\gamma$ is a very short curve in $M(f)$ with non-trivial image in $\pi_1(S^1)$. We must bound the genus of the fibre, $S$, from below. So, isotope $S$ to be a minimal surface. The thick-thin decomposition tells us that there is a large radius embedded tube $N(\gamma)$ about $\gamma$. The fibre $S$ must cross $N(\gamma)$. Thus $S$ has large area and so, because it is minimal, it has large genus. $\Box$

To give the details (and so find actual estimates) you need to understand the geometry of Margulis tubes in hyperbolic three-dimensional manifolds.