The force near a black hole (outside event horizon $r=3r_s/2$) onto a mass $m$ can be calculated by General Relativity:
$$F=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}.$$
However, there must be a distance $r$ where the black hole's gravity basically becomes Newtonian only:
$$F=\frac{GMm}{r^2}.$$
At which distance $r$ does this happen?
From your formula, you can see that $F \approx F_\text{Newtonian}$ when $2GM/c^2r \ll 1 $, or if you rearrange, $2GM/c^2 = r_s \ll r$. In other words, the farther from the Schwarzschild radius, the closer you get to Newtonian gravity.
We can take the ratio $F/F_N$ to get an idea of how far off we are from Newtonian gravity. At $r = 10 r_s$, it is $1/\sqrt{9/10}$ which is about $1.054$, so this is 5.4% off from Newtonian gravity.
At $r = 100 r_s$ it is $1/\sqrt{99/100}$ which is about $1.00504$, so this is 0.504% off from Newtonian gravity.
Although UrsaCalli79's answer is very effective at explaining this mathematically, I will attempt to explain it in layman terms.
You do not leave either Newtonian theory nor Relativistic theory, it is only that one serves better to mathematically predict the forces. This can be seen in the similarity of the equations. Newtonian theory can predict the forces with a fair amount accuracy, and Relativistic theory with more accuracy.
What UrsaCalli79 mathematically demonstrated was that as you take distance from the Schwarzschild radius $-$ which also can be said to be the start of black hole's singularity $-$ the Newtonian equation can more accurately describe the forces.
To end, this essentially means there is a difference of how accurately one theory is versus the other, this difference in accuracy decreases as $r$ increases.
the orbit will be never closed (remember, the system emits gravitational waves), but will spiral towards the center. The smaller $r$ the longer it takes.
The general relativity equations have the term $r_{s}/r$.
When $r_{s}/r \approx 0$, then the equations reduce to Newton's law of gravitation.
A six solar mass black hole has $r_{s} \approx$ 10.875 miles (about 17,400 meters) At 93 million miles, $r_{s}/r \approx 1.1694*10^{-7}$, which makes Newton's equations an extremely good approximation at this distance.
