Let's call an ideal $I\lhd R$  <i>Jacobson</i> if $J(I)=\sqrt{I}$.
I will answer the question in the title by constructing, in stages, a unital ring $R$ where the set of Jacobson ideals
is not a sublattice of the ideal lattice of $R$. <p>

Since the formation of Jacobson radical or nilradical commutes with intersection, it follows that if $A, B\lhd R$ are Jacobson, then $J(A\cap B)=J(A)\cap J(B)=\sqrt{A}\cap \sqrt{B}=\sqrt{A\cap B}$, so $A\cap B$ is also Jacobson. This means that the example I want to construct should have Jacobson ideals $A$ and $B$ where $A+B$ is not Jacobson.<p>

**Stage 1.** Let $\mathbb Q$ be the field of rational numbers and let $L$ be the subring of $\mathbb Q$ consisting of fractions $m/n$ with odd denominator.
The key fact here is that $L$ is a local integral domain with field of fractions equal to $\mathbb Q$. <p>

**Stage 2.**
Let $S$ be the subring of 
$\mathbb Q^{\omega}$ consisting of those tuples
$\textbf{q}=(q_0,q_1,\ldots)\in \mathbb Q^{\omega}$
which are eventually constant and which satisfy the condition that the limit
$q_{\infty}:=\lim_{n\to\infty} q_n$ belongs in the subring $L\leq \mathbb Q$. Saying that $\textbf{q}\in S$
means the same thing as saying that
$\textbf{q}\in \mathbb Q^{\omega}$ and
all but finitely many entries of $\textbf{q}$ are equal to
some fixed $q_{\infty}\in L$.

The key facts here are that<br>

<ul>
<li> $S$ has trivial Jacobson radical. This is because the $n$th
coordinate projection $\pi_n\colon S\to \mathbb Q\colon \textbf{q}\mapsto q_n$
is surjective. Since the image of $\pi_n$ is a field we get that $\ker(\pi_n)\lhd S$ is a maximal ideal of $S$.
This ideal contains exactly those
elements of $S$ that vanish in the $n$th coordinate.
If one intersects the maximal ideals of this type, one is left with
$\{\textbf{0}\}=\{(0,0,\ldots)\}$ only. Thus, if one intersects all maximal ideals of $S$,
one must get the zero ideal of $S$.
</li> 
<li>$S$ has trivial nilradical. This is because the nilradical is contained
in the Jacobson radical.
</li>
<li> $S$ has a retraction onto a
subring $L'$ that is isomorphic to the $L$ from Stage 1.
Here, the retraction is the map $(q_0,q_1,q_2, \ldots)\mapsto (q_{\infty},q_{\infty},q_{\infty},\ldots)$ where $q_{\infty}$ is $\lim_{n\to\infty} q_n$.
</li>
</ul><p>

**Stage 3.** The ring that I have been aiming for is $R:=\{(\textbf{u},\textbf{v})\in S\times S\;|\;u_{\infty}=v_{\infty}\}$. To make this clear, let me repeat the definition using more words than symbols: $R$ is the subring of $S\times S$ consisting of those pairs of tuples that have the same limit.

The key facts here are that<br>

<ul>
<li> $R\leq S\times S$ is a subdirect embedding. That is, $R$ projects onto $S$ in each of the two factors.
This means that the composition of the embedding $R\leq S\times S$
with either projection
$\pi_i\colon S\times S\to S$, $i=1,2$, is surjective.
Let the kernels of these two projections be $A, B\lhd R$, so that
<br>
$A=\ker(\pi_1)=\{\{(\textbf{0},\textbf{v})\in S\times S\;|\;v_{\infty}=0\}$
and
$B=\ker(\pi_2)=\{\{(\textbf{u},\textbf{0})\in S\times S\;|\;u_{\infty}=0\}$. <br>
Notice that any tuple in $A$ or $B$ is zero in all but finitely
many coordinates.
<\li>
<li>
$A$ and $B$ are Jacobson ideals of $R$.
This is because $R/A\cong S$, $R/B\cong S$,
and $S$ has trivial Jacobson radical.
<li>
Let $L''$ be the subring of $R$ consisting of all pairs
$(\textbf{u},\textbf{v})=((\ell,\ell,\ell,\ldots),(\ell,\ell,\ell,\ldots))$ for some $\ell\in L$.
I will refer to $L''$ as the 'diagonal copy' of $L$ inside $R$.
It is clear that $L''\cong L$.
</li>
<li>
Let $C=A+B$. This is the ideal of $R$ consisting of exactly those pairs of tuples in $R$ that are zero almost everywhere.<br> 
I claim that $C$ is the kernel of a retraction of $R$
onto $L''$. To see this, first consider the intersection $C\cap L''$:
since pairs of tuples in $C$ are zero almost everywhere and $L''$
is the diagonal copy of $L$ in $R$, the only common pair
must be the diagonal pair with all coordinates zero. Thus $C\cap L'' = \{(\textbf{0},\textbf{0})\}$.<br>

Now I want to show that $R=C+L''$. Choose any
$x=(\textbf{u},\textbf{v})\in R$
and define $\ell:=u_{\infty}=v_{\infty}$. The pair
$d:=((\ell,\ell,\ldots),(\ell,\ell,\ldots))$ belongs to $L''$
while the difference $x-d\in R$ is zero almost everywhere,
hence belongs to $C$. Thus, $x=(x-d)+d\in C+L''$. This completes the proof that
$C$ is the kernel of a retraction of $R$ onto $L''\cong L$.
</li>
</ul><p>

Now I complete the argument.
We argued that $A, B\lhd R$ are Jacobson ideals.
I must show that $C:=A+B$ is not Jacobson,
i.e., that $J(C)\neq \sqrt{C}$. For this we work in
$R/C\cong L''\cong L$. It suffices to show that
the zero ideal of this quotient is not Jacobson.
Equivalently, we must show that
the $J(0)\neq \sqrt{0}$ in $L$.
But, $L$ is a local domain that is not a field,
so $J(0)\neq 0=\sqrt{0}$. \\\\\\