You actually already proved the difficult part.
If $J$ is maximal, then by the correspondence of ideals you mention $J'$ has to be maximal, too. If it weren't, there would be a proper ideal $T'\supset J'$, which would give a proper ideal $T\supset J$.
Now assume $J$ prime and let $a'b'\in J'$, where $a',b'$ are the classes in $R/I$ of some $a,b\in R$. This means that there are $i_a,i_b\in I$ such that $(a+i_a)(b+i_b)=ab+ai_b+bi_a+i_ai_b\in J+I=J$, as $I\subseteq J$. Thus $ab\in J$ and, say, $a\in J$. Hence $a'\in J'$, so $J'$ is prime.
Finally, suppose $J$ radical and let $(a')^r\in J'$ for some $a\in R$, $r\in\Bbb N$. Then there are some $i_a,i\in I$ such that, by binomial expansion and since $I$ is an ideal, $(a+i_a)^r=a^r+i\in J+I=J$. Hence $a^r\in J$, so $a\in J$. Therefore $a'\in J'$ and $J'$ is radical.
I'll let $N$ represent the intersection of all prime ideals of $R$ which contain $I$. Following through with definitions, you can prove $\sqrt{I}\subset N$. Proving the reverse containinment is not as simple. Given $x\notin \sqrt{I}$, consider the collection $\Omega$ of all ideals $J \supset I$ such that for all $n\in \Bbb N$, $x^n \notin J$. Partially order $\Omega$ by inclusion, and show by Zorn's lemma that $\Omega$ has a maximal element $\frak{p}$. Since $\mathfrak{p}\supset I$, if you can show that $\frak{p}$ is prime, then you can claim $x\notin N$ (since $x\notin \frak{p}$). Take $a,b\notin \frak{p}$, and using maximality of $\mathfrak{p}$, show that $\mathfrak{p} + (ab) \notin \Omega$; this will imply $ab\notin \mathfrak{p}$ and thus $\mathfrak{p}$ is prime.
Best Answer
Yes, it is.
If $I$ is prime, then the conclusion follows easily.
For the converse, write $I=\bigcap_{j\in J} P_j$ with $P_j$ prime ideals. (This can be done since $I$ is radical.) Let's prove that $I$ is prime. Let $ab\in I$ such that $b\notin I$. If $b\notin P_j$ for all $j\in J$, then $a\in I$ and we are done. Otherwise, set $J_1=\{j\in J:b\in P_j\}$. We have $I=I_1\cap I_2$ where $I_1=\bigcap_{j\in J_1} P_j$ and $I_2=\bigcap_{j\in J-J_1} P_j$. Note that $I_1$ and $I_2$ are radical ideals, and moreover $I\subsetneq I_1$ (why?). Since $I$ can not be an intersection of two radical ideals properly containing it, it follows $I=I_2$, so $a\in I$ (why?).