The region is handled Integrate without user having to do any sort of special preparation, other than bounding the region:

Integrate[ ((14 x^2 + 61 x*y + 42 y^2)^3) Boole[-6 <= 2 x + 7 y <= 6 && -6 <= 7 x + 6 y <= 6], {x, -4, 4}, {y, -2, 2}] (* 0 *) 

Or if you can't figure out the bounding box, let Mathematica do it for you:

region = -6 <= 2 x + 7 y <= 6 && -6 <= 7 x + 6 y <= 6; Integrate[ ((14 x^2 + 61 x*y + 42 y^2)^3) Boole[region], {x, MinValue[{x, region}, {x, y}], MaxValue[{x, region}, {x, y}]}, {y, MinValue[{y, region}, {x, y}], MaxValue[{y, region}, {x, y}]}] (*  0  *) 

Update for V10

The region is handled by Integrate without user having to do any sort of special preparation.

Integrate[(14 x^2 + 61 x*y + 42 y^2)^3, {x, y} ∈ ImplicitRegion[-6 <= 2 x + 7 y <= 6 && -6 <= 7 x + 6 y <= 6, {x, y}]] (* 0 *) 

Answer for V9 and earlier (original answer)

Bounding the region by giving appropriate integration limits for x and y allows Integrate to find the integral:

Integrate[ ((14 x^2 + 61 x*y + 42 y^2)^3) Boole[-6 <= 2 x + 7 y <= 6 && -6 <= 7 x + 6 y <= 6], {x, -4, 4}, {y, -2, 2}] (* 0 *) 

Or if you can't figure out the bounding box, let Mathematica do it for you:

region = -6 <= 2 x + 7 y <= 6 && -6 <= 7 x + 6 y <= 6; Integrate[ ((14 x^2 + 61 x*y + 42 y^2)^3) Boole[region], {x, MinValue[{x, region}, {x, y}], MaxValue[{x, region}, {x, y}]}, {y, MinValue[{y, region}, {x, y}], MaxValue[{y, region}, {x, y}]}] (* 0 *)