Ibix said:
No - it's a curvature of spacetime. The distinction is important.
The point about gravity as spacetime curvature is that "things travel in a straight line unless acted on by a force" is no longer true. The correct statement is that "things travel along geodesics unless acted on by a force", and the available geodesics near Earth (or, more precisely, their spatial components) all curve towards the planet and eventually intersect it (unless you can jump upwards at about 11km/s or more). So no force is needed to hold you down - a force is only needed to stop you falling, and that's what the floor provides.
Maybe it's important to stress that these forces are of course not gravitational but mostly electromagnetic.
Concerning the debate, whether gravity is a "force" or not, it's a pretty semantic discussion. It depends on your point of view. Most physicists follow the idea by Einstein that the gravitational interaction is geometrized by General Relativity Theory, i.e., it's the spacetime structure itself that gets dynamical and that gravitational interactions are entirely due to the geometry of spacetime.
Other physicists have a somewhat different point of view and interpret the gravitational interaction as an interaction as all the others, described by a gauge theory, i.e., it is a making a global symmetry local by introducing a gauge field and introducing covariant derivatives via a corresponding connection.
What, however, makes gravitation distinct from the other fundamental interactions (electroweak and strong interactions, described by local gauge theories, where the gauge group acts in an abstract field space) is that what's "gauged" here is indeed the Poincare symmetry of the special-relativistic space-time model (flat Minkowski space), and this pretty formal procedure leads to a theory, which is very closely related to General Relativity. It's extending General Relativity, which assumes that the spacetime model is that of a Lorentzian manifold (i.e., a pseudo-Riemannian manifold with a fundamental form of signature (1,3) or equivalently (3,1) with the uniquely determined metric-compatible connection), to a socalled Einstein-Cartan manifold, where the connection has torsion and is related to the spin of the matter fields. On the macroscopic level, where the only relevant far-distant interaction despite gravity is the electromagnetic interaction, one ends up with a torsion-free connection, i.e., the same spacetime model as GR, and this then can of course immediately reinterpreted as providing the dynamical spacetime metric and thus geometrizing gravity.
One should, however, always be aware of the fact that the corresponding equivalence between inertial forces (i.e., forces occurring due to the use of an accelerating reference frame) is strictly true only locally, i.e., in a sufficiently small space-time region around an arbitrary point in spacetime, where you always can introduce an inertial frame. It's realized by some free-falling non-rotating object. E.g., the Internation Space Station is pretty close to such a local inertial reference frame, and indeed, over a small enough region there are (nearly) no gravitational effects. The astronauts operate in "weightlessness". If, however, you investigate the motion of (extended) bodies within such a free-falling reference frame with more accuracy, you'll realize that the gravitational field is not completely "transformed" away, but that there are still tidal forces left. Mathematically that's pretty obvious: Because at presence of matter spacetime gets a non-vanishing curvature tensor, and this curvature tensor is non-zero independently of the choice of your reference frame (simply because tensors and tensor fields are independent of the choice of a reference frame or coordinates and thus describe observables in GR). So you cannot in any way choose a reference frame, where there are no gravitaional interactions ("tidal forces" cannot be "gauged away").