For a system of point particles, the definition $$\vec{L}_i=\vec{r}_i\times\vec{p}_i$$ is always true; it's just a definition. I see no reason why that won't work here. The only choice you have to make is where to measure the position vectors $\vec{r}_i$ from. A particularly convenient position from which to measure $\vec{r}_i$ is the rotation axis.
One possibly tricky thing here is that the center of mass accelerates. This means there is always a non-zero net force on the system of the two skaters and rope. I suppose it's exerted by the ice on the skaters.
Also, this situation you're describing sounds to me like it does satisfy the requirements of a rigid body, at least for the specific motion you are describing. If the skaters begin to move toward each other so that the distance between them changes, then it wouldn't be a rigid body.
The essence of the question is the definition of the angular velocity and its relation to the axis of rotation.
First of all, to describe the rotation of a single particle, one requires an axis of rotation $\hat{\mathbf{n}}$, and a rate of change of ‘orientation’ (or ‘angular position’) $\frac{d \theta( \hat{\mathbf{n}}) }{d t}$ around that axis. Without these 2 pieces of information, “rotation” is meaningless.
Having the axis of rotation and a rate of change of orientation, one can define the angular velocity $\vec{\omega}$ as
$$
\boldsymbol{\omega} = \frac{d \theta}{d t} \, \hat{\mathbf{n}}
$$
which includes both pieces of information.
Notice that the angular velocity is parallel to the axis of rotation by definition, and has the same amount of ‘information’ as the particle velocity $\mathbf{v}$.
To clarify this further, consider the following figure which depicts the rotation in 3d space, and conforms to the example given in the question,
where the spherical-polar coordinates $\theta$ and $\phi$ are used as the azimuthal and polar angles, respectively and the zenith ($z$-axis) is taken to be parallel to $\hat{\mathbf{n}}$.
In the limit of infinitesimal change in time, $\Delta t \rightarrow 0$,
$$
\Delta \mathbf{r} \approx r \, \sin \phi \, \Delta \theta ~,
$$
and thus,
$$
\left| \frac{d \mathbf{r}}{d t} \right| = r \sin \phi \frac{d \theta}{d t} ~.
$$
One clearly observes that both the magnitude and the direction of $\frac{d \mathbf{r}}{d t}$ (which is perpendicular to the plane defined by the particle position $\mathbf{r}$ and the rotation axis $\hat{\mathbf{n}}$) are given correctly by the cross product
$$
\frac{d \mathbf{r}}{d t} = \hat{\mathbf{n}} \times \mathbf{r} \frac{d \theta}{d t} ~.
$$
Since $\frac{d \mathbf{r}}{d t} \equiv \mathbf{v}$ and $\hat{\mathbf{n}} \frac{d \theta}{d t} \equiv \vec{\omega}$, one obtains
$$
\mathbf{v} = \frac{d \mathbf{r}}{d t} = \boldsymbol{\omega} \times \mathbf{r} ~.
$$
Therefore, the instantaneous axis of rotation $\hat{\mathbf{n}}(t)$ and the instantaneous angular velocity $\vec{\omega}(t)$ are indeed parallel.
Finally, as shown by Gary Godfrey in a comment, one can obtain the correct relation for the angular velocity $\boldsymbol{\omega}$ in terms of the velocity $\mathbf{v}$ and position $\mathbf{r}$ as
\begin{align}
\mathbf{r} \times \mathbf{v} &= \mathbf{r} \times ( \boldsymbol{\omega} \times \mathbf{r} ) \stackrel{\text{BAC-CAB}}{=} \boldsymbol{\omega} \, (\mathbf{r} \cdot \mathbf{r} ) - \mathbf{r} \, (\mathbf{r} \cdot \boldsymbol{\omega}) \\
\Rightarrow \boldsymbol{\omega} &= \frac {(\mathbf{r} \times \mathbf{v}) + \mathbf{r} \, (\mathbf{r} \cdot \boldsymbol{\omega})}{r^2} ~.
\end{align}
This answer is based on Kleppner, D., and R. Kolenkow. “An introduction to mechanics”, (2ed, 2014), pp. 294–295. The figure is taken from the same source.
Best Answer
Cool question.
Let's try to formalize the general concepts presented in the other two responses thus far. Intuitively, the main idea is that the instantaneous axis of rotation is determined by the rotation that takes the object from its configuration at one instant to its configuration at the next instant as opposed to the rotation that takes it from its initial configuration to its configuration at some later time. Let's try to make this precise.
As you already wrote, for any time $t$ we have the following for the position $\vec x(t)$ of a point in the body at time $t$: \begin{align} \vec x(t) = R(t)\vec x(0) \tag{1} \end{align} where $R(t)$ is a rotation for each $t$. Now consider the configuration of the object at a neighboring time $t+\epsilon$. How is the position of a point in the body at this later point related to its position at the point $t$? Well, notice that \begin{align} \vec x(t+\epsilon) &= R(t+\epsilon)\vec x(0) \\ &= \Big(R(t) + \dot R(t)\epsilon\Big)\vec x(0) + O(\epsilon^2) \\ &= \Big(I + \epsilon\dot R(t)R(t)^t\Big)\vec x(t) + O(\epsilon^2) \end{align} For small $\epsilon$, the order $\epsilon^2$ terms become insignificant, so the transformation that takes $\vec x(t)$ to $\vec x(t+\epsilon)$ is the gadget in big parentheses. Let's give it a name: \begin{align} \rho(t, \epsilon) = I+\epsilon\dot R(t)R(r)^t \end{align}
But what exactly is this guy? Well, our intuition tells us that it should be a rotation with axis $\vec\omega(t)$; in particular, $\vec\omega(t)$ should be an eigenvector of this transformation. To see that this is the case,
I make the following:
Claim. $\rho(t,\epsilon)$ takes precisely the form of a "small" rotation about the axis $\vec\omega(t)$. In particular, $\vec \omega(t)$ is an eigenvector of $\rho(t,\epsilon)$.
Proof. First, recall that omega can be defined as the vector that generates the motion of the points in the body as follows: \begin{align} \dot{\vec x}(t) = \vec\omega(t)\times\vec x(t) \tag{2} \end{align} But notice, using eq. $(1)$ that \begin{align} \dot{\vec x}(t) = \dot R(t) \vec x(0) = \dot R(t)R(t)^t\vec x(t) \tag{3} \end{align} Now we notice that the matrix $\dot R(t) R(t)^t$ is antisymmetric since $R(t)$ is an orthogonal matrix for each $t$ (I'll leave this for you to prove. It follows that there exist functions $w_x(t), w_y(t), w_z(t)$ for which \begin{align} \dot R(t)R(t)^t = \begin{pmatrix} 0 & -w_z(t) & w_y(t) \\ w_z(t) & 0 & -w_x(t) \\ -w_y(t) & w_x(t) & 0 \\ \end{pmatrix} \end{align} Using this expression and comparing $(2)$ and $(3)$, it is not hard to show that $w_x,w_y,x_z$ are precisely the components of the angular velocity $\vec\omega$. Putting this all together, we see that $\rho(t,\epsilon)$ can be written as follows: \begin{align} \rho(t,\epsilon) = I + \epsilon\vec\omega(t)\times \end{align} where $\vec\omega(t)\times$ is shorthand for the transformation mapping any vector $\vec v$ to $\vec\omega(t)\times v$. Now, it is easy to show that $\vec\omega(t)$ is an eigenvector of $\rho(t,\epsilon)$; \begin{align} \rho(t,\epsilon)\vec\omega(t) = (I+\epsilon \vec\omega(t)\times)\omega(t) = \vec\omega(t) + \epsilon\vec\omega(t)\times\vec\omega(t) = \vec\omega(t). \end{align} It remains to show that $\rho(t,\epsilon)$ has the form of a "small" rotation about the axis $\vec\omega(t)$. Well, what does such a rotation look like? Well, a general rotation by an angle $\theta$ about an axis defined by a unit vector $\hat n$ can be written as follows (I'll again leave it to you to either show this, or find out why this is true): \begin{align} R(\hat n, \theta) = \exp(\theta\hat n\cdot \vec J) \end{align} Where $\vec J = (J_x, J_y, J_z)$ is a vector of matrices, the so-called rotation generators, whose components are defined by \begin{align} (J_i)_{jk}= -\epsilon_{ijk} \end{align} and $\exp$ is the matrix exponential. Notice that to first order in $\theta$, this shows that \begin{align} R(\hat n, \theta) = I + \theta\hat n\cdot J_i \end{align} But now notice that if we apply this to any vector $\vec v$, then we obtain \begin{align} R(\hat n, \theta)\vec v &= (I+\theta\hat n\cdot \vec J)\vec v \\ &= \vec v + \theta n_i (J_i)_{jk}\vec v_k \hat e_j \\ &= \vec v - \theta n_i\epsilon_{ijk} v_k \hat e_j \\ &= \vec v + \theta \hat n\times \vec v \\ &= (I + \theta\hat n\times )\vec v \end{align} We now see by inspection that $\rho(t,\epsilon)$ has precisely the form of the first order approximation to a rotation $R(\hat n, \theta)$ if we simply take $\hat n = \hat \omega(t)$ and $\theta = \epsilon|\vec\omega(t)|$. $\blacksquare$