I second Willk's answer: gravity doesn't really matter at all for plasma containment, so trying to build a tokamak in orbit would be a huge complication for basically no gain. However, I just took a class on plasma physics so I would be remiss if I didn't cram a bunch more math down peoples' throats.
Now, the main equation that will be governing plasma movement in a tokomak on Earth is
$$m_j n_j \frac{D\mathbf{v_j}}{Dt}=q_j n_j (\mathbf{E+v_j \times B})-\nabla p + m_j n_j \mathbf{g}$$
where $m$ is the mass per particle, $n$ the number density, $q$ the particle charge, $\mathbf{v}$ the fluid velocity, $p$ the pressure, $\mathbf{E}$ the electric field, $\mathbf{B}$ the magnetic field, $\mathbf{g}$ the gravitational field, and the subscript denoting which species we are talking about (normally ion vs electron). Now I know that was a whole bunch to dump at once, but I have a very simple goal here: to show you that that the term involving $\mathbf{g}$ (the gravitational force term) is much smaller than the other forces at play.
You see, the monstrous equation I gave is really nothing more than a dressed up version of Newton's second law: $\mathbf{F} = m \mathbf{a}$. The left hand side is called the convective derivative and decribes how the plasma is being pushed around (analogous to $\mathbf{a}$), while the right hand side lists the forces acting on the plasma. So, let's get a rough sense of the orders of magnitude that we have for the forces.
First off, we will ignore the electric field, since that tends to be approximately zero in steady state plasmas due to a phenomenon called Debye shielding. I'm also going to ignore the term involving the magnetic field because that's the thing we want to adjust.
So, we want to analyze the approximate magnitude of the term $$\nabla p$$
which for thermodynamics reasons is equivalent to
$$\gamma kT \nabla n$$
The plasma recombines at the walls of the vessel, so $n=0$ there. Meanwhile, at the center of a typical fusion plasma we have a typical value of $n=10^{19} m^{-3}$, and a cross sectional radius of maybe $1 m$, giving us an approximate gradient of $10^{19} m^{-4}$. Using the approximation of an isothermal plasma with $\gamma = 1$, and ITER's projected temperature of $kT = 8 keV$, we obtain
$$\nabla p \approx 13 \times 10^3 N/m^3$$
Now, to compare this to the gravitational term. Using the heaviest particle mass (that of tritium ions in the case of ITER) and $n=10^{19} m^{-3}$, we get
$$m n \mathbf{g} \approx 5 \times 10^{-7}N/m^3$$
which is over 10 orders of magnitude less than the force felt due to pressure gradients! So, when you're designing the magnetic field topology, you can pretty safely ignore gravity.
As for an intuitive reason: fusion plasmas are hot. This means particles are bouncing around incredibly quickly, and they bounce off each other so frequently that gravity has basically no time to alter their trajectory in any noticeable way. This is much like how you don't really need to worry about gravity when you're shooting at a target 10 feet away-- the bullet moves so fast that it doesn't really matter.
