Yes.
An open disk in $S^2$ is a set of the form $D(\xi,r) = S^2 \cap B(\xi,r)$, where $\xi \in S^2$ and $r \le 2$. Here $B(\xi,r) = \{\eta \in \mathbb R^3 \mid \lVert \eta - \xi \rVert < r\}$ is the open ball with center $\xi$ and radius $r > 0$. Of course $\lVert - \rVert$ denotes the Euclidean norm.
Note that $D(\xi,r) \subset S^2 \setminus \{-\xi\}$. Equality holds for $r = 2$.
W.l.o.g. we may assume that $\xi$ is the south pole $S = (0,0,-1)$. This is possible because we can find an orthogonal linear automorphism $\phi$ on $\mathbb R^3$ mapping $\xi$ to $S$. By orthgonality we get $\phi(S^2) = S^2$ and $\phi(B(\xi,r)) = B(S,r)$, thus $\phi(D(\xi,r)) = D(S,r)$.
Now take the stereographic projection from the north pole $N = (0,0,1)$
$$s : S^2 \setminus \{N\} \to \mathbb R^2, s(x_1,x_2,x_3) = \frac{1}{1-x_3}(x_1,x_2) .$$
It is well-known that $s$ is a homeomorphism. You can easily check that $\phi(D(S,r))$ is an open disk . It is $= \mathbb R^2$ for $r = 2$; for $r < 2$ just observe that the boundary of $D(S,r)$ has the form $S^2 \cap E(\rho)$, where $E(\rho) = \{ (x_1,x_2, \rho) \mid x_1, x_2 \in \mathbb R\}$ is the hyperplane $x_3 = \rho$ for some $\rho \in (-1,1)$. The set $S^2 \cap E(\rho)$ is a circle in $E(\rho)$ with center $(0,0,\rho)$ and radius $\sqrt{1-\rho^2}$. It is mapped by $s$ to a circle in $\mathbb R^2$ with center $(0,0)$ and radius $\frac{\sqrt{1-\rho^2}}{1- \rho}$. Thus $D(S,r)$ is mapped by $\phi$ to the open disk with center $(0,0)$ and radius $\frac{\sqrt{1-\rho^2}}{1- \rho}$.