I think Question 1 has a positive answer. Denote $B=:B_0(1)$ and $A_r:=\{r<\|x\|<1\} $ for $0<r<1$. For functions $f\in W^{1,2}_0(B)$ we have a Poincaré inequality on $A_r$ :
$$\int_{A_r} f^2dx\le \Big(\frac{1-r}{r}\Big)^2 \int_{A_r} |\nabla f|^2dx \ .$$

Now let $(f_i)$ a sequence in $W^{1,2}(B)$ converging to $f$, so $f^2_i$ converges to $f^2$ in $L^1(B)$. For any sequence $0<\delta_i<1$ we have, by the above inequality:

$$\frac{1}{\delta_i^2}\int_{A_{1-\delta_i}}f_i^2dx\le \frac{1}{\delta_i^2}\int_{A_{1-\delta_i}}f ^2dx + \frac{1}{\delta_i^2} \|f^2-f_i^2\|_{1} \le$$

$$\le \frac{1}{ (1-\delta_i)^2} \int_{A_{1-\delta_i}}\|\nabla f\| ^2dx + \frac{1}{\delta_i^2} \|f^2-f_i^2\|_1\ .$$

Now if $\delta_i=o(1)$, the first term on the RHS is $o(1)$, just because $\|\nabla f\| ^2$ is integrable and $\operatorname{meas}(A_{1-\delta_i})=o(1)$.

On the other hand, if we also choose the sequence $\delta_i$ such that $ \|f^2-f_i^2\|_1=o(\delta_i^2)$ as $i\to\infty$, the other term is also $o(1)$.

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*Details on the above Poincaré inequality.* By density, one can assume $f\in C^\infty_c(B)$. Let $0<r\le \|x\|\le 1$. So $f(x/r)=0$, and we have
$$f(x)= -\int_1^{1/r}\partial_t f(tx)dt=-\int_1^{1/r}\nabla f(tx)\cdot x\ dt\ . $$
Thus by Cauchy-Schwarz
$$ f(x) ^2\le \Big(\frac{1}{r} -1\Big) \int_1^{1/r}\|\nabla f(tx)\|^2 dt\ , $$
and
$$ \int_{A_r} f(x) ^2 \ dx\le \Big(\frac{1}{r} -1\Big) \int_{A_r}\int_1^{1/r}\|\nabla f(tx)\|^2 \ dt\ dx = $$
$$=\Big(\frac{1}{r} -1\Big) \int_1^{1/r} t^{-n} \int_{A_{tr}} \|\nabla f( x)\|^2 \ dx \ dt $$
$$\le \Big(\frac{1}{r} -1\Big)^2 \int_{A_{r}} \|\nabla f( x)\|^2 \ dx \ . $$