Solution to second question in particular case, when $R_0/pR_0$ without zero divisors:
Note that $R_0/pR_0$ is finite dimensional vector space, $\dim R_0/pR_0\leq rk(R_0)$. So easy to see that $R_0/pR_0$ is field. Let for some $x\in R\setminus pR, y\in R, x^2 = py$, let $i\geq0 : r:=p^ix\in R_0, p^{i-1}x\notin R$. From defenition of $r, r\in R_0\setminus pR_0$, $R_0/pR_0$ is field, so $\exists r'\in R_0 : rr'- 1\in pR_0$. So $xrr' = p^ix^2r' = p^{i+1}yr'\in pR$, $x = xrr' - (rr' - 1)x\in pR$, but $x\notin pR$. done
Solution to second question in case $rk(R_0)\leq 2$:
Let $r\in R\setminus pR$, such that $r^2\in pR$. Let $i = min\{i: p^ir\in R_0\}$, $x=p^ir\in R_0\setminus pR_0$. $rk(R_0)\leq 2$, so for some $f = a_0 + a_1t + a_2t^2\in \mathbb{Z}[t], f(x) = 0$, if $p|a_0, a_1, a_2$, we can reduce $f$ by $p$. $a_0 = -a_1x -a_2x^2, a_0^2 = x^2(a_1^2 + 2a_1a_2x + a_2^2x^2)$, so $a_0^2\in p^{2i +1}R$, so $a_0\in p^{i+1}R$, $1/p\notin R$, so $p^{i+1}|a_0$. If $p|a_1$, then $p\nmid a_2$ and $x^2 =(1/a_2)(-a_0 - a_1x)\in pR_0$, $Nil(R_0/pR_0)\not=0$. So $p\nmid a_1$, $a_1x = -a_0 - a_2x^2\in p^{i+1}R$, so $x\in p^{i+1}R, r\in pR$. done
