This is a (rather) expanded version of Yves' answer, because for the specific focus of the question one can understand the proof without getting into many other technicalities of the Borel--Tits paper (which has bigger fish to fry, related to fields of definition and much more). We will explain the proof of the following result:

${\bf Theorem}$ (Borel--Tits): In  a connected linear algebraic group $G$ over an algebraically closed field $k$, every subgroup $H \subset G(k)$ consisting of unipotent elements is contained in a "canonical" connected unipotent subgroup (i.e., one that is stable under all automorphisms of $G$ that carries $H$ onto itself). 


This will use a beautiful construction in section 2 of the Borel--Tits paper. The following is an exposition of some arguments in that paper, with an eye towards the proof of the preceding Theorem (which thereby answers the question posed); by all means one should look at the original paper too (all I have done is provide some more explanation about various aspects of the argument).

Proof: Since the Zariski closure of $H$ is unipotent (but perhaps disconnected), we can replace $H$ with its Zariski closure and instead work with unipotent closed subgroups $U$ in $G$ (e.g., finite ones when in positive characteristic). We induct on the dimension of $G$ (dimension 0 being trivial).  We can also replace $U$ with $U \cdot R_u(G)$ so that $U$ contains $R_u(G)$, and if $R_u(G) \ne 1$ then we can pass to $G/R_u(G)$ to drop the dimension and conclude, so we can assume that $G$ is reductive. We can assume $U \ne 1$, so by nilpotence of unipotent algebraic groups there is a nontrivial central element $u$ in $U(k)$. 

It suffices to show that $u$ lies in the unipotent radical of a "canonical" parabolic subgroup $P$ (i.e., one stable under all automorphisms of $G$ that preserve $u$, or even just preserve the cyclic subgroup generated by $u$).  Indeed, once this is proved, such a $P$ is stable under $U$-conjugation on $G$ (as that leaves the central $u$ in $U(k)$ invariant), so by the self-normalizing property of parabolic subgroups it would follow that $U \subset P$.  But $P \ne G$ (since $R_u(P)$ contains $u \ne 1$ whereas we arranged that $G$ is reductive), so we can use dimension induction to conclude.

We are left with a serious problem concerning a single unipotent element $u \in G(k)$ (with $G$ a connected linear algebraic group): find a "canonical" parabolic subgroup $P$ of $G$ that contains $u$.  (This may not seem like progress, as we have reduced the initial grand claim with canonical parabolics to a problem involving a cyclic group, but we will soon see that this is in fact genuine progress.) The weaker claim that $u$ lies in the unipotent radical of some parabolic subgroup, or equivalently in the unipotent radical of some Borel subgroup, is immediate from Grothendieck's important theorem that $G(k)$ is covered by the subgroups $B(k)$ for Borel subgroups $B$ of $G$ (a result that is proved near the end of section 14 of Borel's textbook but possibly not stated as an official result there; it can be extracted nonetheless). We are going to actually use Grothendieck's theorem to get things started.

By Grothendieck's theorem, $u$ lies in $R_u(B)$, so the cyclic subgroup generated by $u$ is "embeddable" (in the terminology of Borel--Tits), meaning that it does occur inside *some* connected unipotent subgroup (equivalently, in the unipotent radical of a Borel subgroup).  The problem is to show that there is a canonical parabolic subgroup whose unipotent radical contains that cyclic group (or equivalently, contains its unipotent Zariski closure).  Rather than focus on cyclic groups, consider more generally a closed (perhaps disconnected) unipotent subgroup $U$ of $G$ that is "embeddable"; i.e., is contained in a connected unipotent subgroup.  It suffices to
 prove the following refinement of the previous theorem:

${\bf Refined}$ ${\bf Theorem}$ (Borel--Tits). Any embeddable closed unipotent subgroup $U$ of a connected linear algebraic group $G$ is contained in the unipotent radical of a canonically associated parabolic subgroup $P$. (As before, "canonical" means that $P$ is stable under all automorphisms of $G$ that preserve $U$.)

[Note that this result is interesting even when $U$ is connected, and in particular it is interesting in characteristic 0. Also, the embeddable hypothesis is automatic once this is all over!]

Proof:  To construct $P$, we will work with the normalizer $N(U)$, which might be non-unipotent and even more disconnected. Consider the unipotent radical of $N(U)$ (i.e., maximal unipotent connected normal subgroup) anyway.  Note that $R_u(N(U))$ trivially contains $U^0$, but it isn't clear if it contains $U$ or not.  So following Borel--Tits (see 2.4 in their paper), let's put $U$ back in: define the closed subgroup
$$L(U) = U \cdot R_u(N(U))$$
that is unipotent and *does* contain $U$. By the "canonicity" of $L(U)$ in terms of $U$, any automorphism of $G$ that carries $U$ back to itself must do the same for $L(U)$. Also, just as we assumed that $U$ is embeddable (i.e., lies in the unipotent radical of a Borel), we claim the same holds for $L(U)$.  This amounts to checking that $L(U)$ has a fixed point on the projective variety of Borel subgroups of $G$.  By hypothesis $U$ has a fixed point, so the closed (!) locus of Borels containing $U$ is a non-empty projective variety (perhaps disconnected).  Certainly $N(U)$ acts on that locus, so the *connected solvable* $R_u(N(U))$ does too.  By the Borel fixed point theorem, $R_u(N(U))$ has a fixed point in that locus, so we have found a Borel subgroup of $G$ that contains $L(U)$.  Hence, we lose nothing by replacing $U$ with $L(U)$.

We iterate a few times, and eventually reach the situation that the dimension stops growing.  That is, we may assume $L(U)$ has the same dimension as $U$. Thus, the connected normal subgroup $R_u(N(U))$ in $L(U)$ must be contained in $U^0$, so $L(U) = U$.  In other words, we  $R_u(N(U)) \subset U$, so the containment $U^0 \subset R_u(N(U)) = R_u(N(U)^0)$ is an equality. Thus, the normal unipotent subgroup $U$ in $N(U)$ has $R_u(N(U))$ as its identity component.  At this stage, we shall prove that $U$ is already connected (but we really need to go further and get the canonical parabolic, or else the plan of the argument would collapse; in fact, we will show that $N(U)$ is parabolic).

By our arranged "embeddable" hypothesis on $U$, we can pick a Borel subgroup $B$ of $G$ that contains $U$, so $L := B \cap N(U)$ is a solvable (possibly disconnected) closed subgroup of $N(U)$ that contains $U$. The connected solvable $L^0$ must lie in a Borel subgroup $B'$ of $N(U)^0$, and if $B''$ denotes the opposite Borel subgroup then $R_u(B' \cap B'') = R_u(N(U)^0) = U^0$.  We have $B'' = gB'g^{-1}$ for some $g \in N(U)^0$, so $R_u(L^0 \cap gL^0g^{-1}) \subset U^0$.  Equivalently, $R_u(L \cap gLg^{-1}) \subset U^0$. That is, 
$$R_u(N(U) \cap B \cap gBg^{-1}) \subset U^0$$
for some $g \in N(U)^0$ and some Borel $B$ of $G$ containing $U$.  Clearly $U$ is contained in $gBg^{-1}$ too, so 
$$U \subset B \cap gBg^{-1}.$$
But any intersection of two Borel subgroups is always connected (even contains a maximal torus), and unipotent radicals are functorial for *connected* solvable groups, so 
$$U \subset R := R_u(B \cap gBg^{-1}).$$ 
We now show that this inclusion is an equality (so $U$ is connected).

Suppose to the contrary that it is a strict inclusion. Consideration of the descending central series of the connected unipotent $R$ (whose steps must eventually emerge from $U$, maybe at the first step or maybe later) yields a closed connected  subgroup $U'$ in $R$ that normalizes $U$ but is not contained in $U$. This connected unipotent $U'$ then lies inside the identity component of the solvable $N(U) \cap B \cap gBg^{-1}$, so it lies in its unipotent radical, which we have seen is contained in $U^0$. The resulting inclusion $U' \subset U^0$ contradicts that $U'$ is not contained in $U$, so we have shown that $U = R$.  Thus, now $U$ is connected, so we conclude that $U = U^0 = R_u(N(U))$.  

Now it suffices to prove that $N(U)$ is a parabolic in $G$; i.e., we just have to produce a Borel of $G$ inside $N(U)$, where $U = R_u(B \cap B')$ for Borel subgroups $B$ and $B'$ of $G$. We'll show that $B$ is contained in $N(U)$ (by exploiting that also $U = R_u(N(U))$).  Since $U$ contains $R_u(G)$, it is harmless to pass to $G/R_u(G)$ so that $G$ is reductive (if it wasn't already).  We will soon make serious use of the links between reductive groups and root systems (and the uniqueness characterizations of root groups).

There is a maximal torus $T$ of $G$ contained in the smooth connected $B \cap B'$, so $T$ normalizes $R_u(B \cap B') = U$. Thus, $N(U)$ contains $T$. Since $R_u(B)$ is generated by the simple positive root groups for $\Phi(B,T)$ (this is a delicate fact in small positive characteristics), it suffices to show that each of those normalizes $U$.  The $T$-stable connected subgroup $U \subset R_u(B)$ must be generated by the root groups for the nontrivial $T$-weights that occur on its Lie algebra, and we just have to consider simple positive $a \in \Phi(B,T)$ for which the root group $U_a$ is not contained in $U = R_u(B \cap B')$, so $U_a$ is not contained in $B'$.  Hence, $-a \in \Phi(B',T)$ (so $U_{-a} \subset B'$).  

Applying to $B$ the "reflection" associated to the *simple* $a \in \Phi(B,T)$ gives a Borel subgroup $B_a$ containing $T$ and having the same $T$-weights as $B$ except that $a$ is replaced with $-a$.  Thus, $R_u(B \cap B_a)$ is directly spanned by the root groups associated to $\Phi(B,T) - (a) = \Phi(B_a,T) - (-a)$, so it is normalized by $U_{-a}$ (as $-a$ is simple for $\Phi(B_a,T)$). We conclude that $U_{-a}$ normalizes $R_u(B \cap B_a) \cap B'$, so it normalizes its unipotent radical, which is directly spanned by the roots in 
$$(\Phi(B,T) - (a)) \cap \Phi(B',T) = \Phi(B,T) \cap \Phi(B',T) = \Phi(R_u(B \cap B'),T)
= \Phi(U,T).$$
In other words, $R_u(B \cap B_a) \cap B'$ has unipotent radical $U$, so $U_{-a}$ normalizes $U$ yet $U_{-a}$ is not contained in $U^0$ (since $-a$ is not even a $T$-weight for $B$).  Now recall that $U^0 = R_u(N(U)) = R_u(N(U)^0)$, so the $T$-weight $-a$ occurring on $N(U)^0$ does not arise on the unipotent radical, and hence (by reductivity of the quotient  $N(U)^0/R_u(N(U)^0)$ that contains $T$) the $T$-weight $-(-a) = a$ occurs on $N(U)^0$.  This forces $U_a \subset N(U)^0$, so $U_a$ normalizes $U$.  To summarize, for each simple $a \in \Phi(B,T)$, either $U_a \subset U$ or else $U_a$ normalizes $U$, so either way always
$U_a \subset N(U)$.  This forces $B \subset N(U)$, so we are done. QED