1. Since $g(x)=-|x+2|+3$, we have $g(x)\geqslant -2$.

This implies that $|x+2|\leqslant 5$, which further implies that $-5\leqslant x+2\leqslant 5$.

Solving for $x$, we get $-7\leqslant x\leqslant 3$.

Thus, the solution set for the inequality $g(x)\geqslant -2$ is $\\{x|-7\\leqslant x\\leqslant 3\\}$.

2. Since $f(x)=|2x-1|+2$ and $g(x)=-|x+2|+3$, we have $f(x)-g(x)=|2x-1|+|x+2|-1$.

Let $h(x)=|2x-1|+|x+2|-1$. Then,

$$h(x)= \begin{cases} -3x-2,& x\leqslant -2 \\ -x+2,& -2 < x < \dfrac {1}{2} \\ 3x,& x\geqslant \dfrac {1}{2} \end{cases}$$

Thus, $h(x)\geqslant \dfrac {3}{2}$.

Since $f(x)-g(x)\geqslant m+2$ always holds true for $x\in R$, we have $m+2\leqslant \dfrac {3}{2}$.

Solving for $m$, we get $m\leqslant - \dfrac {1}{2}$.

Hence, the range of real number $m$ is $\boxed{(-\infty, - \dfrac {1}{2}]}$.