As the question says, is a connected unipotent subgroup $U$ of a linear algebraic group scheme $G$ always contained in a Borel subgroup of $G$? I have an argument for why the answer is yes, and I have a somewhat confusing example that the answer is no. Obviously one (or both!) of these is wrong but I can't spot the error.
Argument for yes: Unipotent subgroups are solvable and a solvable connected group must be contained in a maximal solvable connected group, i.e., a Borel subgroup.
That seems too simple to be false, so I suspect the problem is with the counterexample which is not so simple.
Counterexample: Let $G = \mathrm{PGL}_2$ over an algebraically closed field $k$ of characteristic $2$. The first Frobenius kernel $G_1 \leq G$ is a closed subgroup whose group ring $kG_1$, which is the dual of the coordinate ring $k[G_1]$, is isomorphic as a Hopf algebra to the universal enveloping algebra of $\mathrm{Lie}(G) = \mathfrak{pgl}_2$. This means any Lie subalgebra of $\mathfrak{pgl}_2$ which consists of nilpotent elements is, by Engel's theorem, a unipotent subgroup of $\mathrm{PGL}_2$. Well, take the subalgebra $$\mathfrak e = \begin{bmatrix}0 & k \\ k & 0\end{bmatrix}.$$ The enveloping algebra of $\mathfrak e$ is $k[x, y]/(x^2, y^2)$. This is self dual so as a group scheme $\mathfrak e$ is connected. This means it should lie in a Borel subgroup of $\mathrm{PGL}_2$, and hence lie in the Lie algebra of that Borel subgroup. But $\mathrm{PGL}_2$ is semisimple of type $A_1$, the unipotent radical of any Borel has a $1$-dimensional Lie algebra so $\mathfrak e$ can't be contained in it.
I assume that you're working over $k$ algebraically closed to avoid existence of Borel issues. If not, let me know, and I can rewrite this.
Let $G$ be a connected linear algebraic group over $k$. Then, recall that a Borel subgroup of $G$ is a maximal connected smooth unipotent subgroup of $G$. In particular, clearly if $U$ is a smooth connected unipotent subgroup of $G$, then $U$ is contained in a Borel. But, if $U$ is not smooth then there is clearly no need. I don't understand your counterexample, but I have a hunch that this is the issue.
Here's a silly example to show that without smoothness hypotheses unipotent subgroups needn't be contained in Borels:
Example: Let $k$ be an algebraically closed field of characteristic $p$. Let $G=\alpha_p$, which is the kernel of the Frobenius map $\mathbb{G}_a\to\mathbb{G}_a$. Then, $\alpha_p$ is certainly unipotent, but the only Borel subgroup of $G$ is the trivial subgroup.