No. The issue is that quasiconformality constants are much more sensitive to local distortion than lengths. This is morally a $L^\infty$ (quasiconformal constants) to $L^1$ (lengths) comparison: $L^\infty$ bounds on a function on a finite measure space imply $L^1$ bounds, but in general the converse fails.
To sketch a counter-example, take a hyperbolic surface with an $\epsilon$-short curve for $\epsilon > 0$ very small. Such a curve has a long collar neighborhood that is close to a standard form. Leaving the core curve fixed, stretch a small annular neighborhood yielding a dilatation of $2$ at the core curve. Compensate for the stretch in a slightly larger neighborhood to produce a $C^1$ homeomorphism supported in a small annular neighborhood $V_\delta$ of the core curve. Call this map from our hyperbolic surface to itself $f_\delta$. By construction, $K(f_\delta) \geq 2$.
One sees that for any geodesic $c$, the length of $f(c)$ is close to the length of $c$. Taking $\epsilon, \delta$ close to $0$, the numbers $\sup\limits_{c} l(f_\delta(c))/l(c)$ and $\inf\limits_c l(f_\delta(c))/l(c)$ can be made to be arbitrarily close to $1$. The key thing to do here is to bound the distortion of geodesic segments in $V_\delta$ in terms of the homotopy class (rel $\partial V_\delta$) of the segment.