The answer is yes.

Since curvature is 1, there is an isometric immersion $\iota\colon D\looparrowright \mathbb{S}^2$. Note that the curve $f(\partial D)$ has constant curvature, therefore $\iota(\partial D)$ bounds a round disc $\Delta\subset\mathbb{S}^2$.

Acually we have two choices for $\Delta$, but for the right choice we get an isometry $D\to \Delta$.
The latter can be done by using Morse-type argument for a function $f\:\Delta\to\mathbb{R}$.