Yes it is possible, but it may not be as efficient as the other methods listed in that PDF file. In fact, I'm still not sure there's any problem for which QUBO is the best way to solve it (see this: What are some real-world applications of QUBO?, and if interested in this area you may also find this one interesting: Are there any real-world problems where quadratization helps to solve something that couldn't have been solved without quadratization?). I'll also say that there's currently an open bounty worth up to 300 points on this question: Where/when did the fields of Operations Research and Spin Physics or Molecular Dynamics begin to cross-pollinate?.
Now for my answer to your question:
Let's start with a simpler case:
Say we have the energy $E$, for the spin Hamiltonian $H = Js_1s_2 + hs_1$ at the configuration $(s_1,s_2) = (1,1)$:
$$
\tag{1}
E = Js_1s_2 + hs_1.
$$
This implies:
$$\tag{2}
E - Js_1s_2 - hs_1 = 0.
$$
Now the following miniminization is equivalent to solving for the unknowns $J$ and $h$ in Eq. 2:
$$\tag{3}
\min_{J,h}\left( E - Js_1s_2 - hs_1 \right)^2,
$$
because the function we're minimizing is a square of a real number, so it cannot be negative, and therefore the minimum cannot be less than 0, and since we know that Eq. 1 is true, we know that the function can be 0 (so the minimum is 0).
Now if we represent $J$ and $h$ in binary, we get:
\begin{align}\tag{4}
J &= J_02^0 + J_12^1 + J_2 2^2 + J_32^3 + \ldots + J_N2^N\\\tag{5}
h &= h_02^0 + h_12^1 + h_2 2^2 + h_32^3 + \ldots + h_N2^N,
\end{align}
where $\left(J_0,J_1,J_2,J_3,\ldots J_N,h_0,h_1,h_2,h_3,\ldots,h_N\right)$ are all either 0 or 1, and then we use:
$$\tag{6}
s_i = 2b_i - 1,
$$
where $b_i$ are the binary variables such as $J_i$ and/or $h_i$, then we have Eqs. 4-5 in terms of $s$ variables (-1,1 rather than 0,1), and Eq. 3 is then in QUBO form!
If we had more than one energy, such as:
\begin{align}\tag{7}
E_1 = Js_1s_2 + hs_1, (s_1,s_2)=(1,1)\\\tag{8}
E_2 = Js_1s_2 + hs_1, (s_1,s_2)=(1,0),
\end{align}
then instead of Eq. 3 we can have:
$$\tag{9}
\min_{J,h}\left( \left(E_1 - J(1)(1) - h(1) \right)^2 + \left( E_2 - J(1)(0) - h(1)\right)^2\right).
$$
This is how myself and others have converted the problem of factoring numbers into QUBO: see Eqs 10 and 11 of "Quantum factorization of 56153 with only 4 qubits" and how they are formed from Eqs 7-9 of that paper.
Now for the "inverse Ising problem" defined in the PDF you gave:
Instead of Eq. 1 we have:
$$\tag{10}
P = \frac{e^{\left(Js_1s_2 + hs_1\right)}}{\sum_{s_1,s_2} e^{\left(Js_1s_2 + hs_1\right)}},
$$
and the same techniques can be used to form the QUBO problem again. If you need more assistance on that, then I suggest you ask another question on how to exponentiate binary variables in QUBO-type problems, since it's a mathematical technique that would be a bit of a tangent to the main thrust of this answer, but would need even more space to explain. It's a bit more complicated than doing the exponential in the case where no actual variables have to be exponentiated, as in the other question you asked here: How to exponentiate binary variables in QUBO-type problems?.
