This question has a negative answer is many respects. Firstly, there are simple constructions in the commutative case. Namely, for $F = K(\!(x)\!)$ we have $F^{\times} = x^{\mathbf{Z}} \times O^{\times}$ where $O = K[\![x]\!]$, so $${\rm{Hom}}(F^{\times},\mathbf{Z}/p\mathbf{Z}) = (\mathbf{Z}/p\mathbf{Z}) \times {\rm{Hom}}(O^{\times}, \mathbf{Z}/p\mathbf{Z})$$ (where "Hom" means "continuous homomorphisms), so to show that the commutative profinite $O^{\times}$ is not topologically finitely generated it is sufficient to show that the left side above is infinite. But by local class field theory the left side is ${\rm{H}}^1(G_F,\mathbf{Z}/p\mathbf{Z})$ (continuous cohomology), and by Artin-Schreier this is $F/\wp(F)$ where $\wp(f) = f^p-f$. Consideration of polar parts in $F = K(\!(x)\!)$ shows that $F/\wp(F)$ is infinite. Thus, the group $K[\![x]\!]^{\times}$ is not topologically finitely generated. Likewise, via the surjection $$\det:{\rm{GL}}_n(K[\![x]\!]) \rightarrow K[\![x]\!]^{\times}$$ the same happens for any $n$.

  But this is really cheating: clearly the "correct" question should get away from the silly constructions that one can make with ease in the commutative setting, so one ought to replace GL$_n$ with semisimple groups $G$ over $O = K[\![x]\!]$. It is also a bad idea to consider general closed subgroups, since with Borel subgroups the torus factor allows us to use the same silly commutative constructions; e.g., if $G = {\rm{SL}}_n$ as an $O$-group with $n > 1$ and $B \subset G$ is the standard Borel $O$-subgroup then $B(O)$ has the $(n-1)$-fold direct product of copies of $O^{\times}$ as a quotient, so obviously $B(O)$ is not topologically finitely generated.

This question has a negative answer is many respects. Firstly, there are simple constructions in the commutative case. Namely, the additive group $K[\![x]\!]$ is an infinite dimensional $\mathbb F_p$-vector space and thus not finitely generated. Less simply, the multiplicative group $K[\![x]\!]^\times$ has infinitely many homomorphisms to $\mathbb F_p$. That is, the quotient by the group of $p$-th powers is infinite dimensional. The group of $p$-th powers is easy to calculate because the $p$-th power is Frobenius, a ring homomorphism: it is the power series of the form $\sum a_n t^{np}$, $a_0\ne0$. Thus the elements $1+t^n$, for $n$ not divisible by $p$, forms a linearly independent sequence in the $\mathbb F_p$-vector space of the mod $p$ quotient. Class field theory relates the slightly bigger group of the mod $p$ quotient of $K((x))^\times$ with the abelianized Galois group of $K((x))$ and with the quotient of the additive group by Artin-Schreier map $f\mapsto f^p-f$.

Thus for any $n$ the whole group $\rm{GL}_n(K[\![x]\!])$ is not finitely generated because it surjects via the determinant to the multiplicative group, which is not. So one might restrict to $\rm{SL}_n$. But this contains subgroups that are not semisimple, such as $\mathbb G_a$, $\mathbb G_m$, $\rm{GL}_{n-1}$, and the Borel subgroup. None of these are finitely generated, but all for the same commutative reason. But this is really cheating: clearly the "correct" question should get away from the silly constructions that one can make with ease in the commutative setting, so one ought to replace GL$_n$ with semisimple groups $G$ over $O = K[\![x]\!]$.