Say a Jacobson ring is one in which each prime ideal is an intersection of maximal ideals.
And also let's call $R$ "rad-nil" if $J(R)=nil(R)$. (This is something I picked up in Lam's *First course in noncommutative rings*, and I don't know if it's used elsewhere, really.)

"Jacobson" indeed implies that rad-nil, since it implies $J(R)\subseteq nil(R)$.

However, the converse isn't true. To show this, just pick a countable local domain which isn't a field (say, $S=\Bbb Q[x]_{(x)}$) and map onto it from $R=\Bbb Q[x_1,x_2,\ldots]$, the polynomial ring in countably many variables. For this homomorphism $f$, we have $R/P\cong S$ where $P=\ker f$ is a prime ideal.

Now $R$ certainly has the property $J(R)=nil(R)$ since $J(R)=\{0\}$.

On the other hand, since $R/P$ is local but not a field, there is only one maximal ideal in $R/P$ containing $P$, so $P$ is not an intersection of maximal ideals of $R$.

So despite $R$ being rad-nil, it is not Jacobson.

The problem is fixed if we say "$R$ is Jacobson iff $R/P$ is rad-nil for every prime ideal $P$ of $R$." This is more or less obvious since "$R/P$ rad-nil" says exactly that $P/P$, the nilradical, is the intersection of $M/P$ where $M$ ranges over all maximal ideals of $R$ lying above $P$, and so $P$ is an intersection of maximal ideals.

Here's the statement that appears in Bourbaki's Algebra Chs 1-7 book in section 3.4 on Jacobson rings visible in googlebooks

Notice they say in Proposition 3 *"every ideal"*:

This is also true since each prime lying over $I$ can be replaced with an intersection of maximal ideals of $R$ over $I$.

So, we have the following nice equivalences:

$R$ is Jacobson

$R/P$ is Jacobson-semisimple (=semiprimitive) for all prime ideals $P\lhd R$

$R$ is "totally rad-nil" (=all quotients are rad-nil)