It is stated that the Chern-Simons action is a topological invariant that is proportional to the Chern-Simons form. But the latter is just a conformal invariant. How do we reconcile these views? Both require integration, so I don't see how the action can be metric-independent but not the CS invariant.
I'll put my thoughts on this question as an answer to reduce the length of the comments. However, this answer should not(!!) be interpreted as a formal answer, but merely as a lengthy comment (there are undoubtedly more qualified people on this site to give a full answer).
First of all, in the case of Chern-Simons theory with a gauge fields in any other (Lie) group then the O$(p,q)$ structure group  of the (orthonormal) frame bundle, it should be immediately clear that the Chern-Simons invariant (so the form itself may transform nontrivially, only the integral should be invariant) is indeed a topological invariant since the only involved quantities are:
- Differential forms (with values in a Lie algebra not related to the metric)
- Exterior algebra
- Integration over manifolds (which is perfectly well-defined without reference to a metric)
Edit (thanks to @mikestone) : The gravitational Chern-Simons theory is not topological as there exists a metric dependency. For example in 3D the variation (under a change of metric) of the Chern-Simons action gives the Cotton tensor. An invariant quantity is only obtained after integrating over the moduli space of metrics.
The information on the following stackexchange page could also be useful: Is gravitational Chern-Simons action "topological" or not?
 I used a general Lorentzian signature just for completeness.
Paper by Jackiw and collaborator: https://arxiv.org/abs/gr-qc/0308071