If the loop is convex then one solution is to use a simple polyhedron clipping algorithm.
First, Generate a non-planar polygon from the path of the loop, it sounds like this is already available. Then for each segment of the polygon define a plane that is perpendicular to the surface of the sphere and passes through the polygon segment. The normal for the plane is tangent to the sphere at that point and contains a point on the line..
Using the plane, iterate over every triangle in the sphere and classify the triangles as being on the negative side of the plane (distance to plane for all vertices of the triangle are negative), the positive side of the plane (all distances are positive) or crossing the plane (some combination of positive and negative).
With the plane normal computed to point inside the polygon:
Discard triangles on the negative side, keep triangles on the positive side, and clip triangles to fit that cross.
Repeat this process for every segment of the polygon.
At least 1 more plane should be created that clips away the back side of the sphere. (usually just choose 3 points on the polygon then form a plane from them, and offset it by a small amount so it doesn't intersect the polygon). Create another plane offset in the opposite direction could be done to complete a convex polyhedron surrounding the region to be clipped.
This is similar to how decals are generated for a polygon mesh in some algorithms.
Another way to compute the planes normal is to compute the vector from the center of the sphere to a point on the polygon segment in question. Then compute a vector for the polygon segment take the cross product and normalize. This will give you a new vector that is perpendicular to the other two. Just make sure the ordering is done correctly so that new vector point to the inside of the loop. The plane distance becomes a point on the segment being clipped.
computing a normal to the plane using the two line segments:
If the sphere center is $r$, the line has end points $a,b$. Then the equation becomes $pN = normalize( cross( a-r, a-b))$ and the point for the plane becomes $a$. To put the plane into the implicit form compute d with $d= dot(-pN, a)$ and the final plane becomes $[pN | d]$. To compute the distance to this plane take the 4d dot product of the point and the plane which will result in a signed distance to the plane. Where the homogenous points is of the form $(x,y,z,1)$
If the loop is not convex then a non-convex, non-planar (for polygon's that do not self intersect) polygon triangulation algorithm could be used. In this method triangulate the polygon, then align all vertices with the surface of the sphere (using a line formed between the center of the sphere and the generated point). Then tesselate to get triangle area near some preset value.