Regarding cases where $G(\mathbb{Z})$ is not maximal in $G(\mathbb{R})$: If I am just allowed to take $G$ to be any simple group scheme over $\mathbb{Z}$, then I take
$$\left\{ (X,Y) : X \left[ \begin{smallmatrix} 2&0 \\ 0&1 \\ \end{smallmatrix} \right] = \left[ \begin{smallmatrix} 1&0 \\ 0&2 \\ \end{smallmatrix} \right] Y,\ \det X = \det Y = 1 \right\} \subset (SL_2)^2.$$
In other words, this is the functor which, to any commutative ring $R$, assigns the set of solutions to these equations in pairs of $2 \times 2$ matrices over $R$.

Over $\mathbb{Z}[1/2]$, this is just isomorphic to $SL_2$, since we can parameterize it as 
$$(X,\ \left[\begin{smallmatrix} 2&0 \\ 0&1 \\ \end{smallmatrix}\right] X \left[\begin{smallmatrix} 1&0 \\ 0&1/2 \\ \end{smallmatrix}\right] ).$$
However, the pair
$$(\left[\begin{smallmatrix} a&b \\ c&d \end{smallmatrix}\right],\ 
\left[\begin{smallmatrix} a&2b \\ c/2&d \end{smallmatrix}\right])$$
is a $\mathbb{Z}$-valued point if and only if $c \equiv 0 \bmod 2$, so $G(\mathbb{Z})$ is the subgroup $\Gamma_0(2)$ of $SL_2(\mathbb{Z})$.

I do not know an example which does not seem artificial like this.