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*][1], 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*][2], Sean Lawton & Adam S. Sikora 
Algebras and Representation Theory volume 20, pages 1133–1141(2017).

Remark: In the above answer, I assumed the OP wanted a *fixed* representation of $G$.  As noted in the answers to this [MO question][3], 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*][4]; it is not exactly the same thing, but it seems to imply the result.


  [1]: https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-168/issue-2/Invariant-theory-of-special-orthogonal-groups/pjm/1102620558.full
  [2]: https://arxiv.org/pdf/1604.02164.pdf
  [3]: https://mathoverflow.net/q/25439/12218
  [4]: https://arxiv.org/pdf/1503.08279.pdf