What is $d\nu$? Do you mean the measure induced by the induced metric on $\gamma$ as a submanifold? In that case the answer is no.
Let $\mathcal{M} = S^1\times S^1$ with coordinates $(x,y)\in [0,2\pi)^2$, with the line element $ds^2 = dx^2 + \epsilon dy^2$. And let $\gamma$ be the geodesic $y = 0$. The induced Riemannian metric on the geodesic is $d\ell^2 = dx^2$.
Let $f = 1$ the constant function. Then $\int_\gamma f d\nu = 2\pi$, while $\int_{\mathcal{M}}fd\mu = 4\pi^2 \epsilon$. By choosing $\epsilon$ sufficiently small you invalidate the proposed inequality.
More generally, purely by dimensional considerations you see that it is unreasonable to expect any expressions of the form you wrote to hold, except possibly in the case that $\gamma$ is treated as a (measure zero) subset of $\mathcal{M}$, and $d\nu$ is the measure $d\mu$, and $f$ is non-negative. Then the LHS is trivially zero and the RHS is non-negative.