Let $p$ be a prime and $n > 1$ a height. My conventions for Morava K-theory are that $K_p(n)^*(pt)=\mathbb{F}_p[v_n,v_n^{-1}]$, $|v_n|$ (the degree of $v_n$) is $2(p^n-1)$.
Question: If $G$ be a connected, compact Lie group, is $K_p(n)^*(BG)$ a Noetherian ring?
As in the paper Local Gorenstein duality in chromatic group cohomology by Pol and Williamson that Drew Heard pointed out, this can be proved by mimicking the proof of the Quillen--Venkov theorem. However as the OP points out, the argument given there is vague regarding convergence issues. Let me discuss how to repeat that argument, somewhat more carefully. Write $E :=K_p(n)$.
We choose $G \to U(n)$ a faithful representation, and consider the homotopy fibre sequence $$U(n)/G \to BG \to BU(n)$$ and its associated Atiyah--Hirzebruch--Serre spectral sequence $$E_2^{s,t} = H^s(BU(n) ; E^t(U(n)/G)) \Rightarrow E^{s+t}(BG).$$ This is a multiplicative spectral sequence, and has a multiplicative map from the Atiyah--Hirzebruch spectral sequence $$F_2^{s,t} = H^s(BU(n) ; E^t(*)) \Rightarrow E^{s+t}(BU(n)).$$ In particular, each $E_r^{*,*}$ is a module over $F_r^{*,*}$. As $H^*(BU(n);\mathbb{Z}) = \mathbb{Z}[c_1, \ldots, c_n]$, the latter spectral sequence is supported in even total degree, so all differentials are zero.
Let me first discuss internal aspects of the spectral sequence, and then convergence aspects.
Using the module structure, each row $E_2^{*, t}$ is a left module over $F_2^{*, 0}=H^*(BU(n);\mathbb{F}_p) = \mathbb{F}_p[c_1, \ldots, c_n]$, and the latter is a noetherian ring. Furthermore each $E_2^{*, t}$ is a finitely-generated $F_2^{*, 0}$-module. The subgroup $Z_\infty^{*,t}/B_1^{*, t} \subset E_2^{*, t}$ of permanent cycles (modulo $d_1$-boundaries) is a sub-$F_2^{*, 0}$-module, so is also a finitely-generated $F_2^{*, 0}$-module: thus its quotient $E_\infty^{*, t}$ is too.
For convergence I refer to Boardman's Conditionally convergent spectral sequences. This kind of spectral sequence is discussed in Theorem 13.12. It is a "half-plane spectral sequences with entering differentials" so converges conditionally to $E^*(BG)$. Now each $E_2^{s,t}$ is a finite-dimensional $\mathbb{F}_p$-vector space, so there can only be finitely-many non-zero differentials out of each such group. It follows that the derived $E_\infty$-page vanishes: $RE_\infty=0$. By Theorem 7.1 in Boardman's paper, the spectral sequence therefore converges strongly. That is, the filtration of $E^*(BG)$ is exhaustive, Hausdorff, and complete, and $$E_\infty^{s,t} \cong F^s E^{s+t}(BG)/F^{s+1} E^{s+t}(BG).$$
The rows $E_\infty^{*, t}$ are finitely-generated $\mathbb{F}_p[c_1, \ldots, c_n]$-modules, so using that $E^*(*)$ is periodic it follows that $E_\infty^{*,*}$ is a finitely-generated $E^*(*)[c_1, \ldots, c_n]$-module. If $x_1, \ldots, x_k$ are module generators, with $x_i \in E_\infty^{s_i, t_i} \cong F^{s_i} E^{s_i+t_i}(BG)/F^{s_i+1} E^{s_i+t_i}(BG)$, we can choose lifts $$\bar{x}_i \in F^{s_i} E^{s_i+t_i}(BG).$$ These, and the $E^*(BU(n)) = E^*(*)[[c_1, \ldots, c_n]]$-module structure, give a map of filtered abelian groups $$E^*(*)[[c_1, \ldots, c_n]]\{\bar{x}_1, \ldots, \bar{x}_k\} \to E^*(BG)$$ which, by construction, is surjective on associated graded.
It is now an exercise (given below) using the definitions of exhaustive, Hausdorff, and complete to see that this map must be surjective, so that $E^*(BG)$ is a finitely-generated $E^*(*)[[c_1, \ldots, c_n]]$-module. In particular it is a noetherian ring.
The exercise (not very elegant): Suppose $f : A \to B$ is a map of filtered abelian groups where $A$ is complete and $B$ is exhaustive and Hausdorff, and $f$ is a surjection on associated graded. Let $b \in B$, so $b \in F^sB$ for some $s$ (as $B$ is exhaustive). Then there is an element $a_s \in F^s A$ such that $f(a_1) \equiv b \mod F^{s+1}B$, so $b - f(a_s) \in F^{s+1}B$ and so there is an element $a_{s+1} \in F^{s+1} A$ such that $b-f(a_s) - f(a_{s+1}) \in F^{s+2} B$, and so on. The sequence $$a^n = a_s + a_{s+1} + \cdots + a_n \in A$$ has, for $n \geq m$, $a^n - a^m \in F^{m+1}A$, so is Cauchy: as $A$ is complete it has a limit $a \in A$, satisfying $a \equiv a_s + a_{s+1} + \cdots + a_n \mod F^{n+1} A$. In particular it satisfies $f(a) \equiv f(a_s) + \cdots + f(a_n) \mod F^{n+1}B$ so $f(a) \equiv b \mod F^{n+1}B$ for all $n$. Thus $f(a)-b \in \cap_{n} F^{n+1} B$, so as $B$ is Hausdorff $f(a)=b$.
