Yes, it can have nonzero curvature, moreover, it will typically have nonzero curvature. Consider, for example, the quotient of $\mathbb{R}^4$ by the standard Hopf $S^1$ action. The quotient will be a cone over a round sphere $S^2$ (it is easy to see from the fact that the quotient of $S^3$ by Hopf action is $S^2$, and standard dilation $x \rightarrow \lambda x$ acts conformally on the metric and commutes with action, you can also calculate it by hands).

Now, such cone only has $0$ curvature if it is actually a Euclidean space. Lets check its parameters. The standard radius $1$ sphere $S^3$ has riemannian $3$-volume $2\pi^2$, the length of every orbit is $2\pi$, which means that the quotient $2$-sphere has the volume $\pi$, which gives the radius $1/2$.

That shows that this cone is not flat - because it has the sphere $S^2$ of radius $1/2$ on the distance $1$ from the origin.

edit: I happened to play a lot with such quotients (also of $S^n$ and some other homogeneous compact spaces), so if you have some additional questions maybe I can answer.

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