The answer is `no', but for a stupid reason: you can have an action with an
ineffective kernel, meaning that (normal closed) subgroup of $G=S^1$
consisting of those $g$ which act trivially: for all $x$ in $M$, $gx = x$. For example,
take a free action of $S^1$ on $M$. Define a new action $g * x = g^p x$
(I'm thinking of the circle multiplicatively). The new action's ineffective kernel consists of the pth roots of unity.

You can get rid of the ineffective kernel by taking $G$ and
dividing by the ineffective kernel. In this circle case, the resulting action of this ```
new'
$S^1$ will be free a.e.: that is it will have principal orbit type the identity (check out, for example, Hsiang's book,
```

Cohomology Theory of Topological Transformation Groups', p. 10-12, esp. p.12) which means, combined with the ```
slice theorem' (p.10) or
the
```

Gleason theorem' (p. 9) the answer to your question becomes `yes'.