This is a comment on the manuscript rather than an answer to your question. When I read that ${\mathbb Z}$-Euclidean principal binary quadratic forms correspond to norm-Euclidean quadratic orders, my immediate reaction was the guess that the non-principal forms must correspond to Lenstra's Euclidean ideal classes. This is in fact true, as a simple calculation shows. The nontrivial ideal class in $K = {\mathbb Q}(\sqrt{-5})$ is generated by the prime ideal ${\mathfrak a} = (2,1+\sqrt{-5})$. By definition, this class is Euclidean if for every $\xi \in K$ there is an $\eta \in {\mathfrak a}$ such that $N(\xi - \eta) < N{\mathfrak a} = 2$. The identitiy $2(2x^2 + 2xy + 3y^2) = (2x+y)^2 + 5y^2$ then shows that this is equivalent to the form $2x^2 + 2xy + 3y^2$ being a Euclidean quadratic form over ${\mathbb Z}$. Thus it seems that this form is missing from Houriet's list, and I believe that this is the only form missing.
The known cases of Euclidean ideal classes for real quadratic fields (disc $K = 40, 60, 85$) show that the answer to problem 2 is negative: not every primitive Euclidean form represents $1$; one example is given by $2x^2 - 5y^2$, the nonprincipal form with discriminant $40$.
This does not answer your actual question, but perhaps it shows that one has to be very careful with making conjectures that seem to be plausible in this area.
Edit. Let $R$ be the ring of integers in a number field $K$, and let $S = R[i]$ denote a subring of the ring of integers in $L = K(i)$. I guess it is easy to see that the form $x^2 + y^2$ is Euclidean over $R$ (here and below: with respect to the absolute value of the norm) if and only if $S$ is Euclidean. But $disc\ S = \pm 4 (disc\ R)^2$ shows that if $S$ has class number $1$, then so does $R$ (because the UFD $S$ necessarily is the full ring of integers, so $disc\ L = \pm 4 (disc\ K)^2$). This shows that if $x^2 + y^2$ is Euclidean over $R$, then $R$ is a PID. Something similar clearly goes through for any binary quadratic form over the integers.
The fact that the proof of this special case already uses class field theory indicates that one should not expect a 3-line proof of the general claim that if $R$ admits a Euclidean quadratic form, then $R$ is a PID.
Edit 2. Take $R = {\mathbb Z}[\sqrt{34}]$ and $q(x,y) = x^2 - (3+\sqrt{34})xy-2y^2$; observe that the discriminant of $q$ is the fundamental unit of $R$, and that its square root generates $L = K(\sqrt{2})$. Then q is Euclidean over $R$ since the ring of integers in $L = K(\sqrt{2})$ is generated over $R$ by the roots of $q$, and since $L$ is Euclidean by results of J.-P. Cerri (see Simachew, A Survey On Euclidean Number Fields). But $R$ is not principal ($L/K$ is an unramified quadratic extensions), so the answer to your question, if I am right, is negative.
