The homotopy type seems to be independent of considering this group with $C^0$ or $C^{\infty}$ topology. As far as I know, the group of oriented diffeomorphisms of $S^2$ is homotopy equivalent to the group $PSL(2, C)$ (or $SO(3, R))$. I do not know direct reference, but maybe I can give a sketch of the proof here:

We are going to prove that the stabilizer of the triple of points $(0, 1, \infty)$ is homotopy equaivalent to a point. First of all, we need to establish the isomorphism between this stabilizer and the space of all smooth almost complex (<=> complex, in 2 dimensions these notions are the same) structures on the sphere $S^2$. In order to do it we use the Riemann's existence theorem, which states, that any complex curve of genus $0$ is isomorphic to $CP_1$:

Step 1: We fix the standard complex structure $I$ on the sphere $S^2$.

Step 2: We map the diffeomorphism $f$ to the structure $f^*I$

Step 3: For $f$ and $g$ if $f^*I = g^*I$ then $g^{-1}f$ preserves $I$ and also $(0, 1, \infty)$ and hence it's identity. So $f = g$. Our mapping is injective.

Step 4: By the Riemann's existence theorem our mapping is surjective.

The space of all almost complex structures on a 2-dimensional sphere is contractable, because it is just a quotient space of the space of all riemannian metrics modulo conformal factors. Only thing we need to check is that nearby almost complex tensors will correspond to the nearby diffeomorphisms, and it can be done using smth like infinite-dimensional inverse function theorem...

And then, weakening the topology to the $C^0$ obviously won't make this space uncontractable.

And on a homeomorphism type... Maybe it makes sense to ask what is the corresponding to the $C^0$ topology on the space of complex structures. I suspect that it's topology given by the norms of quasiconformal mappings, but actually, I dunno.