I would like to explain why the answer to the question is negative in general, following the suggestion of Tom Godwillie (however see PS for a positive answer if $dimX<\infty$)
Let us note first that there is an $S^1$-action on $\mathbb RP^2=M$ such that the quotient is a the segment.
Now, apply the idea of Tom Goodwille. Namely take $\Gamma=\mathbb Z_2$, and consider the inclusion $\mathbb RP^2\to \mathbb RP^{\infty}=K(\mathbb Z_2,1)$ that induces an isomorphism on $\pi_1$. Clearly, this map is not contractible, so it gives a counterexample to the claim in the question.
PS. However, using the remark of Misha it is also possible to prove that the answer to the question is positive in the case when the group $\pi=\Gamma=\pi_1(X)$ has no torsion elements. In particular, it holds when $X$ is finite-dimensional.
Sketch proof. Denote $M/S^1$ by $Y$ and let $p$ be the projection map. I claim that the kernel of the homomorphism $\phi^M_Y:\pi_1(M)\to \pi_1(Y)$ induced by $p$ is generated by torsion elements. Since $\Gamma$ has no torsion elements, it follows that any homomorphism $\phi^M_X:\pi_1(M)\to \Gamma$ factors through a homomorphism $\phi^M_Y$, i.e. there exists $\phi^Y_X:\pi_1(Y)\to \pi_1(X)$ such that $$\phi^M_X=\phi^Y_X\circ\phi^M_Y.$$
Now, since $X$ is aspherical, the homomorphism $\phi^Y_X$ is induced by a map $\psi: Y\to X$. Finally note that the original map $\varphi: M\to X$ and the composed map $\psi\circ p$ are homotopic, since they induce the same homomorphism on $\pi_1(M)$.
