Does the trace distance specify a unique state

For qubits, in the Bloch ball representation: $$rho=frac{1}{2}(I+mathbf{r}cdotmathbf{sigma})$$ $$rho'=frac{1}{2}(I+mathbf{r}'cdotmathbf{sigma})$$ The difference $rho-rho'$ is $$rho-rho'=frac{1}{2}[(mathbf{r}-mathbf{r}')cdotmathbf{sigma}]$$ The eigenvalues of $rho-rho'$ are $$lambda_{pm}=pmfrac{1}{2}sqrt{(x-x')^2+(y-y')^2+(z-z')^2}=pmfrac{||mathbf{r}-mathbf{r}'||_2}{2}$$. Thus the trace distance in terms of Bloch vectors is $$D(rho,rho')=frac{||mathbf{r}-mathbf{r}'||_2}{2}$$. In this picture, $D(rho_i,sigma)$ is the the radius of a ball containing $sigma$ as a point and centered in $rho_i$.

I can think two answers to your question: if you give me a table of values $d_i$ for the six projectors of the Pauli basis I would say that no, $sigma$ is not specified uniquely, since without further restrictions if I give you $d_i=epsilon$ for all $i$ there is $epsilon$ such the intersection between the six balls is empty (or any finite number of projectors). Now if you are asking whether a quantum state $sigma$ can be expressed uniquely via a table of values $d_i$ I would believe so, since a point in $mathbb{R}^3$can be expressed uniquely as the intersection between six spheres (maybe even less).

For higher dimensions in principle one could use similar arguments in terms of qudit Bloch vectors - https://arxiv.org/abs/0806.1174 . (Maybe this relates to the construction of Mutually Unbiased Bases somehow?)