I think it should be possible to show that your distance matrix is always nonsingular by showing that it is always a Euclidean distance matrix (in the usual sense of the term) for a non-degenerate set of points. I don't give a full proof but sketch some ideas that I think can be fleshed out into a proof.
Two relevant papers on Euclidean distance matrices are Discussion of a Set of Points in Terms of Their Mutual Distances by Young and Householder and Metric Spaces and Positive Definite Functions by Schoenberg. They show that an $n\times n$ matrix $A$ is a Euclidean distance matrix if and only if $x^\top Ax\le0$ for all $x$ with $e^\top x=0$ (where $e$ is the vector with $1$ in each component) and that the affine dimension of the points is $n$ if and only if the inequality is strict.
It follows that a Euclidean distance matrix can only be singular if the affine dimension of the points is less than $n$: If the affine dimension is $n$, there cannot be an eigenvalue $0$, since there is a positive eigenvalue (since $e^\top Ae\gt0$), and the span of these two eigenspaces would non-trivially intersect the space $e^\top x=0$, contradicting the negative definiteness of $A$ on that space.
To use all this for your case, one could try to show that a distance matrix in your sense is always a Euclidean distance matrix in the usual sense for points with affine dimension $n$. I think this could be done by continuously varying the exponent $\alpha$ in $A_{ij}=d(x_i,x_j)^\alpha$ from $1$ to $2$ and showing a) that there is always a direction in which the points can move such that $A$ remains their distance matrix with the changing exponent and b) that this movement necessarily causes them to have affine dimension $n$.
To get a feel for how this might work, consider a square: The movement would bend the square into a tetrahedron. The proof would need to account for the fact that this seems to hold only for $\alpha\lt2$; you can see from the example of three points in a line that they can be bent to accommodate $\alpha\lt2$ but not $\alpha\gt2$.
Intro
I'll present an incomplete interpretation framework.
Let's being by considering two points in one dimension. We can easily generalize the results to more than one dimension.
Our matrix is given by,
$$M=\begin{bmatrix} 0 & |x_1-x_2| \\ |x_1-x_2| & 0 \end{bmatrix}$$
The eigenvalues are given by
$$\lambda_1=-\lambda_2=|x_1-x_2|$$
The eigenvectors are given by
$$v_1=\begin{bmatrix} 1 \\ 1 \end{bmatrix} \quad v_2=\begin{bmatrix} 1 \\ -1 \end{bmatrix}$$
There are three parts to providing an interpretation of the above. First we need to know what $M$ does. Second, we need to be aware of what eigenvalues and eigenvectors are.
Interpretation
What does $M$ do? Well, the best way to figure this out is to express it in terms of the things we do know. Comparing this to a reflection matrix, we see that our matrix reflects vector components. In other words, the components of the vectors are switched. In addition, the matrix also scales the vector by the distance between the data points.
So our matrix $M$ operates on a vector $v$. What is $v$? Since it wasn't put in the question, I've taken the liberty of choosing the meaning of $v$. The vector $v$ denotes points the number of trips taken for each data point. As example, consider,
$$v=\begin{bmatrix} n_1 \\ n_2 \end{bmatrix}$$
We have $n_1$ to denote the number of trips between $x_1$ and the other data points. $n_2$ denotes the number of trips taken between $x_2$ to other data points. If we interpret $x_i$ to be actual points in a physical space, then $M(v)=d$ is a vector whose components $d_i$ denote the total distance traveled from $x_i$ to other points in space.
Example
Here's an explicit example. Imagine that you're a busy young adult thinking about where to live to juggle how trips between work, your mom's house, and a friend's house. Work is located at $x_1=55$, Mom's at $x_2=0$, and your friend's house is at $x_3=30$. We have,
$$M=\begin{bmatrix} 0 & 55 & 25 \\ 55 & 0 & 30 \\ 25 & 30 & 0 \end{bmatrix}$$
So lets imagine that you want to decide where to live. These are the only places you ever go, so you really want to pick a place close to one of the locations, effectively right next to one of them if possible. You know that you have to go to and from work 14 times a week wherever you live. You also know that mom's going to want to see you twice a week so you'll be making 4 trips to and from her place. Finally, you're in a band with your friend so you need three practice times a week so that's another six to and from trips. You also know that you'll always have to stop by where you live to get ready to go back out. Work clothes aren't good for band practice or dinner with mom, lugging a guitar around isn't convenient, and mom will be furious if you leave for work or to see your friend. Putting this together we have,
$$v=\begin{bmatrix} 14 \\ 4 \\ 6 \end{bmatrix}$$
So to get the total distance you'll have travel if you live at each of these locations, we need to multiply $M$ by $v$.
$$M \cdot v=d=\begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}=\begin{bmatrix} 370 \\ 950 \\ 470 \end{bmatrix}$$
So if you live next to work, you'll have to travel $d_1=370$ units per week.
If you live next to mom, you'll have to travel $d_2=950$ units per week.
If you live next to your friends, you'll have to travel $d_3=470$ units per week.
So clearly, its best to live next to work. Note that this is very similar to what happens in real life!
However, it may also be important to find $v$ such that,
$$M \cdot v=\lambda \cdot v$$
For instance, imagine the scenario above with a slight modification. Instead of wondering where to live, our young adult had a less hectic job that let him make his own schedule, something more reasonable than working 7 days a week! In this case, they might want to make a number of visits to each location so that the total distance traveled, based at any location, is directly proportional to the number of trips you want to make. Sounds a little idealistic, however, its great for weekly planning.
Generalization
Generally speaking, the situation I talked about was a weighted graph. Each edge was weighted with the distance between vertices. Each vertex represented a location. The Distance Matrix, with applications, is simply the adjacency matrix for the graph with weights included.
Using this we can define new situation using graphs. In fact there are also others posts dealing with meaning of adjacency matrix eigenvalues. For instance, see here.
Best Answer
Well, had to search on the topic for a while. Something concrete had been found in an old, but frequently cited alticle.
Theorem 6 there states:
...where $D$ is a EDM (Euclidean distance matrix) – real symmetric $n \times n$ matrix with elements $-\frac{1}{2}d_{ij}^2$ and with zero diagonal; the dimensionality of $D$ is the dimension of the space containing the points that generate $D$; further details can be found in the original article. (Gower, J. C., Properties of Euclidean and non-Euclidean distance matrices, Linear Algebra Appl. 67, 81-97 (1985). ZBL0569.15016.)
So, seemingly Wikipedia was not precise enough – general position in the case means "general" in terms of hypersphere. In other words, $rank(D) = r + 1$, if and only if the points generating $D$ lie on the relative boundary of an $r$-dimensional hypersphere.