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 $\pi_1(M)$.