The Jacobson-Morozov theorem that any nilpotent element $e$ in the Lie algebra of a simple algebraic group $G$ can be embedded in an $\mathfrak{sl}_2$-triple, has a restriction (in terms of the Coxeter number) on the characteristic of the underlying field (assumed to be algebraically closed). This restriction is also required for the "uniqueness" of the triple, up to $C_G(e)$-action. (This result is due to Kostant.) In his 1980 paper, Pommerening had removed the restriction on the characteristic in Jacobson-Morozov's theorem, up to very small exceptions (i.e., characteristic is "bad"). Does the uniqueness as in Kostant's result also hold with this weaker restriction? If it does, then where does Jacobson-Morozov along with uniqueness result of Kostant fail in positive characteristics?
Jacobson-Morozov – Failure of Jacobson-Morozov in Positive Characteristics
algebraic-groupslie-algebrasmodular-lie-algebrasrt.representation-theory
Related Solutions
Much of the literature in characteristic 0 is older and may not immediately fit the exact format of your question. But here is a sample:
N. Jacobson's 1962 book Lie Algebras (later reprinted by Dover) discusses in II.5 (and II.11) the Lie algebra structure of a completely reducible linear Lie algebra in characteristic 0. His Theorem 8 shows that the Lie algebra decomposes into its center plus a semisimple ideal. (In the parallel algebraic group setting, the Lie algebra of a completely reducible linear algebraic group in characteristic 0 has this form.) The semisimple or just simple case is really the crucial one for Jacobson-Morozov theory.
In the 1968-69 IAS seminar volume published as Springer LN 131 in 1970, look at section III.4 in Conjugacy classes by T.A. Springer and R. Steinberg; here there are also adaptations to prime characteristic.
R.W. Carter's 1985 book Finite Groups of Lie Type (Chapter 5) has a nice treatment, though he usually works with simple algebraic groups.
There is also a Lie algebra treatment in section 11 of Bourbaki's Chap. 8 in the Lie groups series, supplemented by interesting exercises.
In characteristic 0 the exponential map works well to pass from the Lie algebra to the group, but in characteristic $p$ the Jacobson-Morozov argument only works for large enough $p$. For refinements involving the groups when $p$ is small, there are substantial papers by G. McNinch and D. Testerman in the past decade or so. At any rate, the case $g=1$ or $n=0$ of your question is trivial and can be left aside.
As BCnrd observes, you have to be careful to specify your maps to be algebraic group homomorphisms. In characteristic 0, working with an algebraically closed field isn't so important, but in general you have to treat points of the group over a field with care. (There is a careful study by Borel and Tits of the way abstract group homomorphisms relate to algebraic group morphisms, but you don't want to get into that here.)
The short answer is no. In prime characteristic, the Killing form sometimes behaves badly even for simple Lie algebras. If "semisimple" means that the solvable radical is zero, there is no way to obtain the classical equivalences with non-degeneracy of the Killing form and with the direct sum decomposition into simples. Moreover, the simple Lie algebras have only recently been classified when $p=5$ (Premet-Strade), while for $p=2,3$ little is known and for $p>5$ the classification takes an enormous amount of work (by Block-Wilson and others). Conditions on the dimension of a faithful representation or on the prime are not enough to sort out the concept of semisimplicity. Still, a lot is known. For example, Seligman and others explored in the 1960s the class of modular Lie algebras for which the Killing form is non-degenerate.
ADDED: Much more could be said along these lines, but for older results see the book Modular Lie Algebras by G.B. Seligman (Springer, 1967). Modular Lie algebras have been of much less importance overall than linear algebraic groups, whose Lie algebras are "restricted" (have a nice $p$th power operation) but still don't reflect precisely the group structure or representation theory. Moreover, the representations of simple or more generally semisimple Lie algebras in prime characteristic are poorly understood even for those arising from Lie algebras of semisimple algebraic groups. So working with a fixed $p$ and fixed $n$ will usually not be illuminating.
Best Answer
The uniqueness can break down very badly in positive characteristic. Supose $G=SL_p$ where $p$ is the characteristic of the base field. Take a regular nilpotent element $e$ in $\mathfrak{g}=\mathfrak{sl}_p$. Then there is a nilpotent element $f\in\mathfrak{g}$ such that $e$, $f$ and $h=[e,f]$ form an $\mathfrak{sl}_2$-triple with the property that $h^p=h$. Note that the identity matrix $I$ is in $\mathfrak{g}$. It is easy to see that there is $f_0\in\mathfrak{g}$ such that $[e,f_0]=I$ (many lecturers find this fact useful when explaining that Lie's theorem can fail in characteristic $p$). Let $\lambda$ be a scalar such that $\lambda^p\ne \lambda$. Since $h$ commutes with $I$ and $ad\ h$ is semisimple, we may assume further that $[h,f_0]=-2f_0$. Then $(e,h+\lambda I, f+\lambda f_0)$ is another $\mathfrak{sl}_2$-triple containing $e$. If the spans $\mathfrak{s}_1$ and $\mathfrak{s}_2$ of the triples are conjugate under $G$, then restricting the $p$-dimensional vector representaion of $\mathfrak{sl}_p$ to $\mathfrak{s}_1$ and $\mathfrak{s}_2$ we would get equivalent representations of $\mathfrak{sl}_2$. However, the representation we get from $\mathfrak{s}_1$ is restricted whereas the one we get from $\mathfrak{s}_2$ is not. So the triples are not conjugate under $C_G(e)$. One can replicate this example inside any Lie algebra of a reductive group $\widetilde{G}$ whch contains $G$ as a closed subgroup.