Assuming you've placed your two pebbles far from any other masses the only force they will feel is their mutual gravitational attraction. Suppose you place them one metre apart, then the force they feel is given by your equation, and since in your example $m_1$, $m_2$ and $r$ are all equal to one the force each pebble will feel is just $G$ or $6.673 \times 10^{-11}$ Newtons. Each pebble will therefore start accelerating towards the other pebble at $6.673 \times 10^{-11}$ ms$^{-2}$.
Lets say you've made your pebbles from granite, which has a density of about 2700 kg/m$^3$, so the radius of the (spherical) pebbles is 0.045 m. When the pebbles collide, after a bit of bouncing around they'll settle down and remain in contact at a spacing of 0.09m. At this spacing the force between them will be about $8.2 \times 10^{-9}$ Newtons.
You could make the pebbles orbit each other. The general equations for two bodies orbiting each other are a bit complex if you're not up to speed with calculus, but if you're happy with a circular orbit it's easy to calculate the orbital velocity. For an object moving in a circular orbit the acceleration towards the centre is simply $v^2/r$. In the example above we calculated the acceleration to be $6.673 \times 10^{-11}$ ms$^{-2}$, and the radius of the orbit is half the spacing so $r = 0.5m$. That means the orbital velocity is given by:
$$ 6.673 \times 10^{-11} = \frac{v^2}{0.5} $$
so:
$$ v = 5.78 \times 10^{-6} \space \text{m/sec} $$
So if you placed the pebbles a metre apart and set each one moving at 5.78 microns per second they would follow circular orbits about their centre of mass.
If you want to know more about how two bodies orbit each other have a look at the Wikipedia article on the two body problem.