If you want to find the change in length of a single curve $\gamma$, the answer is of course trivial $$\int \sqrt{g(\gamma',\gamma')} \mathrm{d} \gamma \to \int e^\sigma \sqrt{g(\gamma',\gamma')} \mathrm{d} \gamma $$ The problem is that a general conformal transformation changes the geodesic structure. This can be easily verified e.g. by taking a flat Euclidean metric, seeing that scalar curvature becomes non-zero under a conformal transform and the Levi-Civita connection is thus suddenly non-integrable (i.e. the geodesics are qualitatively different geodesics).

I am more of a dimension 4 person, but I am quite certain that a general $\sigma$ has no closed formula for a change in distances even in the all so special dimension 2.

There might be some luck for *harmonic* $\sigma: \nabla^i \partial_i \sigma=0$, because for those in dimension 2 the Ricci tensor is invariant and the scalar curvature only rescales by an $e^{-2 \sigma}$ factor. This means that e.g. in the case of a flat Euclidean metric the result is again flat because apart from Ricci tensor and it's trace, the Riemann curvature tensor has only the Weyl-tensor degree of freedom which is invariant with respect to conformal transformations.