Equivalent measures let $\mu,\sigma$ be finite measures on the complex unit circle $\mathbb{T}$. Would it be correct to say that $\mu\sim\sigma$ implies that $L^2(\mu,\mathbb{T})=L^2(\sigma,\mathbb{T})$ as topological spaces?

I would not say that $L^2(\mu)=L^2(\sigma)$ as topological spaces, meaning that they are homeomorphic, since any two separable Hilbert spaces are. So this would be a trivial claim. 
I would rather say that there exists a natural Hilbert space isomorphism between $L^2(\sigma)$ and $L^2(\mu)$: indeed, by Radon-Nikodym's theorem, there exist measurable $f, g\ge 0$ such that $d\sigma=f d \mu$ and $d\mu=gd\sigma$, so that $fg=1$ both $\sigma$- and $\mu$-almost everywhere. Then the isometric mappings 
\begin{align}
L^2(\sigma)\to L^2(\mu),& & L^2(\mu)\to L^2(\sigma) \\
u \mapsto u \sqrt{f},& & v \mapsto v \sqrt{g}
\end{align}
are inverse to each other and so they are Hilbert space isomorphisms.
