Here is my partial progress, from a while ago. I don't think I really got anywhere.

Let $I$ be the sheaf of ideals that vanish on the image of the diagonal morphism. Then we are required to find a sheaf $J \subset I$ locally generated by sections such that $J=I$ on some open neighborhood of the diagonal. We can just take $J$ to be the maximal subsheaf in $I$ locally generated by global sections. We need to choose some neighborhood of the diagonal. I cannot see any process to choose a neighborhood other than the union of $U_0 \times U_0$ for all $U_0$ affine open. Given a section, we want to extend it to all affine open neighborhoods of $U_0 \times U_0$. Since it is enough to check on a basis of affine opens, it is enough to check that it extends to all affine opens of the form $A \times B$ for $A,B \subset X$ affine open with $U_0 \subset A \cap B$.

Thus we have a scheme $A \cup B$ for $A,B$ affine and an affine open $U_0 \subset A \cap B$. We have a function on $U_0 \times U_0$ that vanishes on the diagonal. Without loss of generality this is $f \otimes 1 - 1 \otimes f$ for $f \in U_0$. We want to extend this to a function on $A \otimes B$. Such a function could take the form $r \otimes u - s \otimes t$ where $r$ and $s$ are functions on $A$, $t$ and $u$ are functions on $B$, and $s$ and $u$ are nonvanishing on $U_0$, $f=r/s=t/u$ on $U_0$ and $ru-st=0$ on all of $U$. It is easy to find pairs satisfying every condition but the last. But there is no obvious way to get them to satisfy the last condition.

One approach would be to first find $r/s$ satisfying the first two properties, then extend $r$ and $s$ from $A \cap B$ to $B$ as fractions, then divide those fractions as $t$ and $u$. Unfortunately, you can only write each function as a fraction on each affine - on different affines they could be different fractions, so $ru-st$ need not be $0$. It is easy to create examples of unextendable functions

But it is hard to produce a counterexample from this, because if you make the function unextendable on $B$, an adversary trying to prove that the morphism is a strong immersion can just try to extend it the same way on $A$. Thus, you need to find an affine scheme with a very bad open subset, and then embed that subset into a very different affine scheme where it is still very bad but in a different way.

My attempts to forcibly construct a counterexample led to a big confusing mess. You want to set up a "dictionary" where nice well-behaved functions on $A$ become very ugly glued-together messes of fractions of functions $B$, with the functions in those fractions discussed in a second dictionary where they are turned into ugly messes on $A$, so someone who tries to extend a function consistently both ways is left with an infinite regress. But it is not clear that defining this does not create an infinite regress, or if it doesn't what the properties of the resulting object are.