 Solution

No. If the image lies in a great circle, there is a fixed unit vector $a$ with $⟨N,a⟩=0$, so $a$ is tangent to $S$ everywhere. Since the ambient derivative of this constant vector is zero, the shape operator satisfies $S(a)=0$. One principal curvature is therefore zero. Minimality makes their sum zero, so both vanish and $S=0$. The Gauss map is then locally constant, and connectedness makes it constant on all of $S$, contradicting its image being a great circle.