I think CGAL i too robust for such simple task. And Sutherland Hodgman clipping is usefull, when you have general polygon. But you have axis aligned box and you should make use of it.
So my opinion:
First case: Triangle and box are intersecting. You can figure it out by line-line intersection test (for example this: http://mathworld.wolfram.com/Line-LineIntersection.html but you can google it in many ways - also look at EDIT2 at the end of this answer). You make line from every edge of box and every edge of triangle. Then you find intersection of all combinations (there are 12 cases <- 3 (triangle) x 4 (box) edges).
If you find intersection point (IP), test, if IP lies between two vertices of triangle or somewhere else on line (created from triangle edge). If IP lies between vertices, then test, if IP's x and y coordinates lie in box x and y intervals (it's axis aligned) - it's green case on picture. If IP doesn't lie in both of it's intervals, then it's red case on picture.
After this test, you should have several vertices, which define your new polygon created from triangle.
//EDIT: Oh. I forget, you have to take also every vertex from triangle for your new polygon, which lies in box (on picture there are two such vertices).
If no edge-edge intersection is found, you have to check two other cases:
Second case: Whole rectangle lies in triangle - your new points from triangle are just vertices of box.
Third case: Whole triangle is in box. Take original triangle as result.
I hope, this can work. It's not tested. Maybe you can use some heuristic and avoid of testing all possible edge-edge combinations.
//EDIT2: You should also make use of axis aligned box in line-line intersection test - just find y coordinate with given x constant (in case of vertical box edge) or find x coordinate with given y constant (in case of horizontal edge).