Understanding the total spin as the Noether's charge and rotation generator of the Heisenberg model Consider the Heisenberg model where the Hamiltonian $$H= J\sum_{\langle i,j\rangle}\textbf{s}_i\cdot \textbf{s}_j$$ has continuous rotational symmetry. Since $\textbf{s}_i\in\mathbb{R}^3$, the rotation matrices as acting on each $\textbf{s}_i$ be represented by a matrix $$R(\hat{\textbf{n}},\theta)=\exp{[i(\textbf{J}\cdot\hat{\textbf{n}})\theta]}\tag{1}$$
where $R(\hat{\textbf{n}},\theta)\in SO(3)$ and $\textbf{J}=(J_1,J_2,J_3)$ are the three-dimensional representation of $SO(3)$ generators. It is trivial to show that the Heisenberg hamiltonian is invariant under (2) by using $\textbf{s}_i\to R\textbf{s}_i$ and using $R^TR={\rm identity}$.
However, section 1 of this note claims that the Noether's charge is given by the total spin $\textbf{S}=\sum\limits_{i}\textbf{s}_i$, and $[\textbf{S},H]=0$. Therefore, the rotation is represented by $$U(\boldsymbol{\theta})=\exp{[i(\textbf{S}\cdot\boldsymbol{\theta})/\hbar]}.\tag{2}$$ 

Questions
$\bullet$ Firstly, as far as I know, the Heisenberg model (like the Ising model) is also a model of classical moments. In that case, where does $\hbar$ come in the rotation matrix?
$\bullet$ It is not clear how $\textbf{J}=\textbf{S}$ or whether $\textbf{J}$ hasany relation with $\textbf{S}=\sum\limits_{i}\textbf{s}_i$. 
 A: *

*From the $so(3)$ Lie algebra 
$$ [\hat{s}^a_j,\hat{s}^b_k]~=~i\hbar\delta_{jk}\sum_{c=1}^3\epsilon^{abc}\hat{s}^c_k ,\tag{1}$$
it follows that $$\hat{\bf S} ~=~\sum_j \hat{\bf s}_j\tag{2}$$ generates rotations.

*Normalize $$\hat{\bf s}_j \leadsto\frac{\hat{\bf s}_j}{\hbar}\tag{3}$$ to get a Lie algebra (1) that survives the classical limit $\hbar\to 0$.

*It is overkill to use Noether's theorem. It is just Heisenberg's EOM $$\frac{d\hat{\bf S}}{dt}~=~\frac{1}{i\hbar}[\hat{\bf S},\hat{H}]~=~\hat{0}.\tag{4}$$
A: In the quantum system, $S = \sum_k s_k$ are the total angular momentum operators. It is easy to check that indeed $[H,S]=0$, so $S$ are conserved charges and generate a symmetry. These generators act on the local spin operators as follows (I'm setting $\hbar=1$ everywhere).
\begin{align}
[S_a, s_{i,b}] = \epsilon_{abc} s_{i,c} = (J_a)_{bc} s_{i,c} \,.
\end{align}
Here $i$ is the site index, and $a,b=1,2,3$ is the spin index.
$J_a$ is the $SO(3)$ generator, a $3 \times 3$ matrix.
Notice that the action of $S_a$ on the local operators is the same as the action of the $SO(3)$ generator $J_a$, but $S_a$ acts on the Hilbert space components while $J_a$ acts on the spin index. We say that $\vec{S}$ is a linear representation of the rotation algebra $so(3)$ on the Hilbert space. This is the sense in which $\vec{S}$ and $\vec{J}$ are related: $\vec{J}$ defines an `abstract' algebra, and $\vec{S}$ is a representation of this algebra on our physical Hilbert space.
Given a representation of the algebra $so(3)$ in terms of Hermitian operators $\vec{S}$, we can construct a unitary representation of the group $SO(3)$ by writing your eq. (2). These $U(\theta)$ are unitary operators on the Hilbert space. Their action on the local spins is given by (up to signs of $\theta$)
\begin{align}
U^\dagger(\theta) s_{i,a} U(\theta) = R(\theta)_{ab} s_{i,b} \,.
\end{align}
Namely, these unitary operators mix the three operators of $s_{i,a}$, $a=1,2,3$ in the same way that $SO(3)$ does when we think of $\vec{s}_i$ as a vector. 
You also asked why $\hbar$ shows up in the finite group elements and how this changes in classical systems. SRS Qmechanic gave a nice answer for how to take the classical limit.
Let me also explain how things work in a typical classical Hamiltonian system, although I'm not sure this story applies to the Heisenberg model.
The simple exponentiation of generators in (2) to give finite group elements works for linear representations (e.g. for representations on a Hilbert space) like we have in the quantum system. In the classical system the rules are different. Instead of commutators we use Poisson brackets. In order to get a finite group element we need to exponentiate something like $\{S,\cdot\}_{\mathrm{p.b.}}$, and recall that the Poisson brackets involve derivatives of the form $\frac{\partial}{\partial x} \frac{\partial}{\partial p}$ with coordinates $x$ and momenta $p$. These derivates have dimension of $\hbar^{-1}$ so putting them in instead of $\hbar^{-1}$ keeps the dimension correct. See for instance this review for more details.
