Here's another example.
Consider $\mathbb{R}P^3$ in the model of a 3-ball with antipodal boundary points identified.
Consider the $G = \mathbb{Z}/2\mathbb{Z}$ action on $\mathbb{R}P^3$ given by the antipodal map sending $(x,y,z)$ to $-(x,y,z)$.
I claim that $\mathbb{R}P^3/G$ is homeomorphic to a cone on $\mathbb{R}P^2$. To see this, use coordinates on $C\mathbb{R}P^2$ given by $([x,y,z],t)$ where we think of $(x,y,z)\in\mathbb{R}^3$ and we're collapsing $\mathbb{R}P^2\times\{0\}$ to a point. Now,map $(x,y,z)$ in $\mathbb{R}P^3$ to $\big([x,y,z], (x^2+y^2+z^2)\big)$ (and map the origin to the cone point). This is clearly continuous away from the origin. It's not too hard to see that it's continuous at the origin as well.
It's also not hard to see that this descends to a bijective map from $\mathbb{R}P^3/G$ to $C\mathbb{R}P^2$, which is therefore a continuous bijection between compact Hausdorff spaces, so is itself a homeomorphism.
Finally, notice that $C\mathbb{R}P^2$ is not a topological manifold (with or without boundary) because of the cone point $p$. A neighborhood $U$ of the cone point $p$ has, by excision, $H_k(U, U-p)\cong H_k(C\mathbb{R}P^2, C\mathbb{R}P^2-p) = H_{k-1}(\mathbb{R}P^2)$, which means $p$ can be neither a manifold point nor a manifold-with-boundary boundary point.
The answer is positive even in greater generality:
Suppose that $G$ is a locally compact group with (at most) countably many connected components, $M$ a locally compact Hausdorff topological space, $G\times M\to M$ a continuous action. Let $p\in M$ be a point whose $G$-orbit $Gp$ is closed in $M$. Then the orbit map induces a homeomorphism
$$
G/G_p \to Gp\subset M,
$$
where $Gp$ is equipped with the subspace topology. Here $G_p$ is the stabilizer of $p$ in $G$. This result is Theorem 2.13 in the book by D.Montgomery and L.Zippin "Topological Transformation Groups''.
Note that Theorem 2.13 is stated in the case of transitive group actions. However, in general, the action of $G$ on its orbit is transitive and the assumption that the orbit is closed implies that the orbit is a locally compact space (since we are assuming that $M$ is).
Note that for this to apply you have to assume that Lie groups are, say, 2nd countable (some people do not make this assumption), so that they have (at most) countably many components. Otherwise, the result is false.
Best Answer
A nice instructive example is to take the group $\mathbb{Z}$ and let it act on $S^1$ by an irrational rotation. That is, let $\alpha\in[0,1]$ be some irrational number and define $n\cdot e^{i\theta}=e^{i(\theta+2\pi n\alpha)}$ for $n\in\mathbb{Z}$. The action is free since $\alpha$ is irrational, so $e^{2\pi i n\alpha}\neq 1$ for any nonzero integer $n$. It is not proper because $S^1$ is compact but $\mathbb{Z}$ is not.
Note that since the fractional parts of integer multiples of $\theta$ are dense in $[0,1]$, every orbit of this action is dense in $S^1$. So the orbit space is quite horrible: it has the indiscrete topology!