In general, the answer to the original question is **no**.

In fact, $S^1$ acts on $\mathbf{RP}^2$ in such a way that that the action has a fixed point. The quotient space is homeomorphic to a closed interval, hence $\pi_1(\mathbf{RP}^2/S^1)=\{1\}$ whereas $\pi_1(\mathbf{RP}^2) = \mathbf{Z} / 2 \mathbf{Z}$.

The description of the action is the following. We see $\mathbf{RP}^2$ as the quotient of $S^2$ by the antipodal map $x \mapsto -x$. Then the action of $S^1$ on $S^2$ given by rotations around the vertical axis descends to $\mathbf{RP}^2$. All the points have trivial stabilizer, except the point corresponding to the class of the two poles (which is fixed) and the points corresponding to classes on the equatorial circle (whose stabilizers have order $2$).

Hovewer, the answer is **yes** when $M$ is *simply connected*. This is a consequence of a more general result of Armstrong concerning actions of compact Lie groups on simply connected spaces, see the paper

M-A. Armstrong, *Calculating the fundamental group of an orbit space*, Proceedings of the American Mathematical Society **84** (1982), 267-271, in particular Example 4.