We are given $f(x)+f(\pi-x)=\pi^2 \quad \forall x\in\mathbb{R}.$ We need to evaluate $I=\int_0^{\pi} f(x)\sin x\,dx.$

Use the substitution $x\mapsto \pi-x.$ Let $I=\int_0^{\pi} f(x)\sin x\,dx.$ Now put $x=\pi-t$, so $dx=-dt$. When $x=0$, $t=\pi$; when $x=\pi$, $t=0$. Thus, $I=\int_{\pi}^{0} f(\pi-t)\sin(\pi-t)(-dt)=\int_0^{\pi} f(\pi-t)\sin t\,dt.$ Renaming $t$ as $x$, $I=\int_0^{\pi} f(\pi-x)\sin x\,dx.$

Add the two expressions for $I$: $2I=\int_0^{\pi}\big(f(x)+f(\pi-x)\big)\sin x\,dx.$ Using the given condition, $f(x)+f(\pi-x)=\pi^2,$ so $2I=\int_0^{\pi} \pi^2\sin x\,dx=\pi^2\int_0^{\pi}\sin x\,dx.$

Evaluate the sine integral: $\int_0^{\pi}\sin x\,dx=[-\cos x]_0^{\pi}=(-\cos\pi)-(-\cos0)=1-(-1?)$ More carefully, $[-\cos x]_0^{\pi}=-\cos\pi-(-\cos0)= -(-1)-(-1)=1+1=2.$ Hence, $2I=\pi^2\cdot 2=2\pi^2.$ Therefore, $I=\pi^2.$

So the correct option is $\boxed{\text{A: }\pi^2}.$