You can define the "force of gravity" in a static space-time as the proper acceleration of a stationary particle. Though you might have to define a preferred coordinate system to define the notion of a static particle. The issue I'd be concerned with is eliminating circular orbits as not being a "stationary particle".
You can also write this force as the gradient of a scalar potential. See Wald pg 158 problem 4. However, it won't follow Gauss's law, the surface integral of the "force" around an enclosing body won't be constant. However, it turns out (see Wald around pg 288)that if you do a Gauss-law like intergal on the "force at infinity" rather than the force, you do get a constant quantity that's proportional to the enclosed Komar mass.
You can also look at
http://en.wikipedia.org/wiki/Komar_mass though Wald is a better source and the inspiration for this calculation.
The force at infinity is just the force mulltiplied by the redshift factor. The redshift factor can be defined as \sqrt{\xi^a \xi_a} where \xi^a is the timelike Killing vector of the static system. Using sensible coordinates, which you probably need to define the notion of a stationary particle anyway, the timelike Killing vector will just be a unit vector [1,0,0,0] and the redshift factor is equal to \sqrt g_{tt}, i.e. the square root of the time dilation factor.
For the Schwarzschild metric, the effective potential (in geometric units) whose gradient yields the force is ln(g_tt), the effective potential whose gradient yields the force-at-infinity is sqrt(g_tt).
As handy as all this is for static space-times, you can't generalize it - in general, there's no such thing as gravitational potential in GR.