I propose the following: suppose $\,U\,$ is not prime, thus there exist $$\,x,y\in R \text{ such that }x,y\notin U\,,\,xy\in U\,.$$ Define now $B:=U+\langle y\rangle$.
By maximality of $\,U\,$ we have that $\,B\,$ is f.g., say $$\,B=\Bigl\langle u_i+r_iy\,,\,1\leq i\leq k\,,\,k\in\mathbb{N}\,\,;\,\,u_i\in U\,,\,r_i\in R\Bigr\rangle$$ and let now $$U_y:=\{s\in R\,\,;\,\,sy\in U\}.$$ (1) Check that $\,U_y\,$ is a proper ideal in $\,R\,$.
(2) Show that $\,U_y\,$ is f.g.
Put $\,U_y=\langle s_1,\ldots,s_m\rangle\,$, and take $\,u\in U\Longrightarrow\,\exists v_1,\ldots,v_k, t_1,\ldots,t_k\in R\,\,s.t.$$$u=\sum_{n=1}^kv_nu_n+\sum_{n=1}^kt_nr_ny.$$
(3) Show that $\displaystyle{\sum_{n=1}^kt_nr_n}\in U_y.$
(4) Putting $\,\Omega:=\{u_i,\ldots,u_k,ys_1,\ldots,ys_m\}\,$, derive the contradiction $\,U=\langle\Omega\rangle$.
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.
Best Answer
Here’s a proof that doesn’t involve the quotient $R/A$.
Suppose that $A$ is not prime; then there are $a,b\in R\setminus A$ such that $ab\in A$. Let $B$ be the ideal generated by $A \cup \{a\}$; $B = \{x+ar: x\in A\text{ and }r\in R\}$. Clearly $A \subsetneq B$, so $B = R$, $1_R \in B$, and hence $1_R = x + ar$ for some $x\in A$ and $r\in R$. Then $$b = b1_R = b(x+ar) = bx + bar.$$ But $bx \in bA \subseteq RA = A$, and $bar \in Ar \subseteq AR = A$, so $b \in A$. This contradiction shows that $A$ is prime.