$\mathbf{Q1}$
There are 2 kinematic constraints for this system, using your notation,
$$dx=R\sin\phi\, d\theta,\tag{1}\label{1}$$
$$dy=-R\cos\phi\,d\theta,\tag{2}\label{2}$$
which become, after divided by $dt$ on both sides,
$$\dot x=R\sin\phi\;\dot\theta,$$
$$\dot y=R\cos\phi\;\dot\theta,$$
and you can use $v=R\,\dot\theta$ to write the constraint equations more compactly. The reason Eq.\eqref{1},\eqref{2} are called nonholonomic constraints is that both of them cannot be reduced to a definite functional form $$F(x,y,\phi,\theta)=0\tag{3}\label{3}$$ to which we refer as geometric constraint. The distinguishing property of Eq.\eqref{3} is, if it exists, that it imposes restriction on the possible values of the generalized coordinates $x,y,\phi$ and $\theta$. Mathematically speaking, Eq.\eqref{3}, if it exists, represents a 3-dimensional surface in the 4-dimensional configuration space of the rolling disk just like $x^2+y^2+z^2=1$ represents a 2-sphere in the 3-dimensional Euclidean space, which we know, prevent arbitrary values of $x,y$ and $z$ to be taken. This means although kinematic constraints \eqref{1} and \eqref{2} exist, they impose no restriction on the values of the 4 generalized coordinates and we can make the disk to move from a given starting position to any other position, i.e. you can even bring the disk from $(x_0,y_0,\phi_0,\theta_0)$ to $(x_0,y_0,\phi_0,\theta_1)$ where $\theta_0\neq\theta_1$ and this clearly violates Eq.\eqref{3} since it assigns a (unique) $\theta$ to a certain set $\{x,y,\phi\}$.
$\mathbf{Q2}$
It is not that $F(x.y,\phi,\theta)\neq0$ is related to the nonholonomicity of a system, it is the non-existence of Eq.\eqref{3} from the kinematic constraints prescribed by the nature of the system.
$\mathbf{Q3}$
@Eli has shown how this is a holonomic constraint mathematically. However, there might be concerns about the periodicity of the generalized coordinates and the configuration. In this question, we are only looking at the constraints without considering the dynamics, so gravity is ignored as usual. Let call the smaller disk of radius $r$ as $A$ and the bigger disk of radius $R$ as $B$. So you might wonder: Is the configuration of the system is the same after $A$ has rolled around $B$ a full revolution ($\theta$ from $0$ to $2\pi$)?
The answer is no. The reason lies in the fact that $\phi$ is not an integral multiple of $2\pi$. That means picking a representative point in $A$ and it has different positions before and after the revolution, therefore the configuration of the whole system is different. But there are 2 cases of interest: whether the ratio $r/R$ is a rational or an irrational number. Suppose $r/R=h$, then after one revolution around $B$, $\phi$ becomes $\frac{2\pi}{h}$. In order to have the same configuration as the original one at the very beginning, we require that when $\theta$ becomes $a(2\pi)$, where $a$ is an integer, $\phi$ must be $b(2\pi)$, where $b$ is an integer as well so that we can make the identification for $\theta=0$ to $\theta=a(2\pi)$ as well as $\phi=0$ to $\phi=b(2\pi)$. If $h$ is a rational number, namely $\frac{c}{d}$ where $c\, \&\, d$ are integers, $\phi$ becomes $\frac{2\pi d}{c}$ after one revolution around $B$ and that means $A$ just needs to rotate around $B$ for $c$ revolutions to have $\phi$ as an integral multiple of $2\pi$. If $h$ is an irrational number, however, the configuration of the whole system will never be the same, no matter how many revolutions $A$ rotates around $B$. There is no periodicity for the $2$ generalized coordinates in this case.
$\mathbf{Update}$
I have recently read a book that gives a rigorous (in the sense of a physicist) mathematical proof that the constraints \eqref{1} and \eqref{2} are not integrable, that is Analytical Mechanics by Nivaldo A.Lemos. The proof utilizes the Frobenius Integrability Theorem and requires the reader to be familiar with differential forms as prerequisite. However, most of the essential materials are included in Appendix B of his book. If you are mathematically oriented, it might be something you are after.
