For surface $G(x,y,z)=0$ which is bounded in $z$ direction on $(x,y)\in[0,1]\times[0,1]$, by specifying large enough $z$-interval, ContourPlot3D
with the option RegionFunction
can be used to generate the required region graphics.
Take $G(x,y,z)=x^3+y^3+z^3-20$ and $F(x,y,z)=30-(x-2)^4-(y-0.5)^4-0.8 (z-0.5)^4$ as an example.
graph = ContourPlot3D[
x^3 + y^3 + z^3 - 20 == 0,
{x, -3, 3}, {y, -3, 3}, {z, -5, 5},
RegionFunction -> Function[{x, y, z},
30 - (x - 2)^4 - (y - .5)^4 - .8 (z - .5)^4 < 1],
BoundaryStyle -> None, Mesh -> None,
PlotPoints -> 50]

Then convert the 3D graphics to 2D one:
Graphics[{##}[[1]],
PlotRange -> All, Frame -> True, Axes -> False,
Sequence @@ Rest[{##}]] & @@
DeleteCases[graph /.
GraphicsComplex[pts_, others__] :>
GraphicsComplex[pts[[All, 1 ;; 2]], others] /.
Polygon[pts_] :>
Sequence[FaceForm[{Lighter[Blue, .8]}], EdgeForm[{Lighter[Blue, .8]}], Polygon[pts]],
_?(MatchQ[#,
(VertexNormals -> _) | (BoxRatios -> _) | (PlotRangePadding -> _) | (PlotRange -> _)
] &), \[Infinity]]

Note: for an unbounded $G(x,y,z)$, it will be much harder to determine whether a point $(x,y)$ satisfies the region condition, and I suppose, as Artes said, impossible to find a general way.
G[x,y,z] == 0
algebraic or transcendental? $\endgroup$ – J. M.'s ennui♦ Aug 21 '12 at 10:21G
andF
. You can find some related ideas from the answers to this question : mathematica.stackexchange.com/questions/8536/… $\endgroup$ – Artes Aug 21 '12 at 10:45