For some Lie groups you will need additional invariants. For example, for the even orthogonals you will need the Pfaffian in addition to the Trace.
See for example:
Invariant theory of special orthogonal groups, Helmer Aslaksen, Eng-Chye Tan, Chen-bo Zhu Pacific J. Math. 168(2): 207-215 (1995).
However, you can already see this with one copy of $\mathfrak{so}(2,\mathbb{F})$ and $\mathrm{SO}(2,\mathbb{F})$ acting by conjugation. For in that case the Lie algebra is $\left\{\left(\begin{array}{cc}0&a\\-a&0\end{array}\right)|\ a\in \mathbb{F}\right\}$ and the conjugation action is trivial. In this case the trace is identically 0 and the Pfaffian is the only useful invariant.
In case you are interested, the Lie group version of this question was addressed by myself and Sikora here:
Varieties of Characters, Sean Lawton & Adam S. Sikora Algebras and Representation Theory volume 20, pages 1133–1141(2017).
Remark 1: In the above answer, I assumed the OP wanted a fixed representation of $G$. As noted in the answers to this MO question, if you vary over all representations, then for $k=1$ the answer is yes. But for $k\geq 2$ the answer appears to me to still be no. See Theorem 1 in Sikora's paper SO(2n,C)-character varieties are not varieties of characters; it is not exactly the same thing, but it seems to imply the result (see the comments for a strategy to fill in the details).
Remark 2: As noted already by Professor Procesi, his work cited by the OP implies the answer is yes for the Lie groups $\mathrm{GL}_n$, $\mathrm{O}_n$, and $\mathrm{Sp}_{2n}$ (by restricting $k$-tuples of generic matrices to the subvariety $\mathfrak{g}^k$). From this one can also deduce that the answer is yes for the Lie groups $\mathrm{SL}_n$ and $\mathrm{SO}_{2n+1}$.
Remark 3: As I said in the comments, I believe the work of G. Schwarz probably implies the answer is also yes for a representation of $G_2$ (he also addresses some Spin groups). The question for the other exceptional groups is open as far as I know. If I were to guess, I would say it is probably true for all of them except $E_6$ which has additional symmetry as in the case of $\mathrm{SO}_{2n}$ (where the answer is no as I already indicated).
Remark 4: Once one knows the answer for a given $G$, then one also knows it for finite central quotients of $G$ (which is how one goes from the orthogonal case to the special orthogonal case for odd $n$). Also, if one knows the answer for $G$ and $H$, then one also knows it for $G\times H$. Putting these observations together with the known cases reduces the entire problem down to the (simply connected forms of the) exceptional groups and the spin groups.