I will humbly disagree with the solution of my fellow trying to justify the opposite answer.
[OP] After a prolonged period of time, is the colour composition uniform?
Under the sole condition of diffusion, a solution made up of several substances can never achieve a uniform composition even at $t \rightarrow \infty$. Moreover, gravity does not influence the composition profile, it only ensures that the fluid does not "fall" by specifying an adequate pressure at every point.
We will consider for simplicity the diffusion evolution along the $z$-axis, so we can imagine that the $x$ and $y$ directions are of infinite lenght than that of the $z$ axis. In this scenario, the velocity field can only have the $z$ direction, and since air doesn't flow in the $x$ or $y$ direction, is only a function of $z$, i.e. $\mathbf{v}(z) = v_z(z)\mathbf{\hat{z}}$. The image is the following:
1. Mass Balance The continuity equation is
\begin{align}
\frac{\partial \rho}{\partial t} + \mathbf{\nabla} \cdot \mathbf{v} = 0 \tag{1}
\end{align}
but a prolonged period of time traduces into equating the time derivative to zero, that is, we want the solutions for $t \rightarrow \infty $ or the steady-state solutions. Eq. (1) says
\begin{align}
\mathbf{\nabla} \cdot \mathbf{v} = 0 \rightarrow
\boxed{\frac{\mathrm{d}v_z}{\mathrm{d}z} = 0} \tag{2}
\end{align}
Thus, the velocity field at $t \rightarrow \infty $ is just a constant for every point. Here we need to make a decision:
- If we consider $v_z \neq 0$, then we cannot stop at $z = L$, because the air will continuously go up. This can be circumvented by considering an ideal solution, of semi-infinite diffusion, where $L \rightarrow \infty$. This is an interesting solution often studied in transport phenomena but not our case.
- If we want the tube of a finite size then $v_z = 0$, which is the condition that we will take.
2. Momentum Balance The equation of motion in the case for a newtonian fluid is
\begin{align}
\rho \left(\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v} \cdot
\mathbf{\nabla}\mathbf{v}\right) &= -\mathbf{\nabla}P + \mu \nabla^2\mathbf{v} + \rho \mathbf{g} \tag{3} \\
\end{align}
where $\mu$ is the viscosity of the air+bromine solution. In a prolonged period of time, employing Eq. (2), and remembering that the gravity points down in our coordinate system
\begin{align}
\rho(v_z \mathbf{\hat{z}}) \cdot \left(\underbrace{\frac{\mathrm{d}v_z}{\mathrm{d}z}}_{=0}\mathbf{\hat{z}}\right)
&= -\frac{\mathrm{d}P}{\mathrm{d}z}\mathbf{\hat{z}} + \mu \underbrace{\frac{\mathrm{d}^2 v_z}{\mathrm{d}z^2}}_{=0}\mathbf{\hat{z}} +
\rho (-g\mathbf{\hat{z}}) \\
0 &= -\left(\frac{\mathrm{d}P}{\mathrm{d}z} + \rho g\right)\mathbf{\hat{z}} \rightarrow
\boxed{\frac{\mathrm{d}P}{\mathrm{d}z} = -\rho g} \tag{4} \\
\end{align}
Eq. (4) establishes the pressure profile in order to balance the weight of the fluid, that can be solved once stating a boundary condition. This is typically the pressure at a boundary, for example, if open to the atmosphere that would be $P(z=L) = P_\text{atm}$. This profile is a linear function and has its maximum at the base of the tube.
3. Mass Balance for Bromine The continuity equation for $\ce{Br2}$ under the absence of sources or sinks (like chemical reactions), and considering Fick's law to relate the flux and the concentration, is
$$
\frac{\partial C_\ce{Br2}}{\partial t} + \mathbf{v} \cdot
\mathbf{\nabla}C_\ce{Br2} = D_\ce{Br2,air} \mathbf{\nabla}^2 C_\ce{Br2}
\tag{5}
$$
where $D_\ce{Br2,air}$ is the diffusion coefficient of $\ce{Br2}$ in air. Again, for steady state conditions Eq. (5) yields
\begin{align}
(v_z \mathbf{\hat{z}}) \cdot \left(\frac{\mathrm{d}C_\ce{Br2}}{\mathrm{d}z}\mathbf{\hat{z}}\right)
&= D_\ce{Br2,air} \frac{\mathrm{d}^2 C_\ce{Br2}}{\mathrm{d}z^2} \\
v_z \frac{\mathrm{d}C_\ce{Br2}}{\mathrm{d}z}
&= D_\ce{Br2,air} \frac{\mathrm{d}^2 C_\ce{Br2}}{\mathrm{d}z^2} \rightarrow
\boxed{\frac{\mathrm{d}^2 C_\ce{Br2}}{\mathrm{d}z^2} - \frac{v_z}{D_\ce{Br2,air}} \frac{\mathrm{d}C_\ce{Br2}}{\mathrm{d}z} = 0} \tag{6}
\end{align}
Eq. (6) establishes the concentration of bromine, for $t \rightarrow \infty$, which is independent of gravity, once two boundary conditions are stated. For our case of $v_z = 0$, Eq. (6) states a linear profile, with a maximum at the base of the tube, and a minimum at the top of the tube. Therefore we have proven that the concentration of $\ce{Br2}$ is not uniform in the tube.
4. Conclusion The lower the diffusion coefficient, the more difference of concentration there will be between the bottom and the top. This is the reason why we mix coffee with a spoon every morning, why we use a magnetic stirrer for titration, for the variables of interest to be independent of position. From a mathematical point of view, those actions are done to avoid this type of behaviour, where we have an undesired dependency of position in the concentration of a substance.
If $\ce{Br2}$ diffuses very easily in air, then we may consider that $C_\ce{Br2} \approx \text{constant}$ in the tube, because we may not be able to notice. However, in this scenario that we are studying, and strictly prohibiting to shake the tube, a difference of concentration and hence of colour is inevitable.
