Is my thinking correct?
Let's suppose there are two independent random complex vectors, $a,b\in\mathbb{C}^N$. Additionally, both vectors have a zero mean, and their covariance matrices are defined (e.g., $\mathbb{E}[a a^H]$).
In this case, can I say $\mathbb{E}[a^H b b^H a] = \mathbb{E}_a\big[a^H \mathbb{E}_b[b b^H] a\big]$? Here, $(\cdot)^H$ is the Hermitian (conjugate transpose), and $\mathbb{E}_a, \mathbb{E}_b$ are the expectation over $a$ and $b$, respectively.
(I have edited my original answer according to comments and to properly address OP's question)
You are correct in this case. Using the law of total expectation we can easily show that
$$\mathbb{E}\big[a^Hbb^Ha\big]=\mathbb{E}_b\Big[\mathbb{E}[a^Hbb^Ha\big|b]\Big]=\mathbb{E}_a\Big[a^H\mathbb{E}[bb^H]a\Big]\,.$$
Now, I want to use this opportunity to clarify why and in which (other) cases this property applies.
The law of total expectation allows us to state, for any function $g(X,Y)$, that $$\mathbb{E}\big[g(X,Y)\big]=\mathbb{E}_Y\Big[\mathbb{E}\big[g(X,Y)\big|Y\big]\Big]=\mathbb{E}_X\Big[\mathbb{E}\big[g(X,Y)\big|X\big]\Big]$$
Consider now a function $g(X,Y)$ such that is linear in one of its variables, e.g. $Y$. Then, if $X$ and $Y$ are independent, thanks to linearity of expectation we can factor out the linear term in the inner conditional expectation, essentially "replacing a variable with its expectation" without worry.
In the case of OP's question, we can write $g(a,B)=a^HBa$ where $B=bb^H$. Therefore, $a$ and $B$ are independent, and $g(a,B)$ is linear in $B$. Thus, applying the law of total expectation, $$\mathbb{E}\big[g(a,B)\big]=\mathbb{E}_a\Big[\mathbb{E}\big[g(a,B)|a\big]\Big]=\mathbb{E}_a\Big[\mathbb{E}\big[a^HBa|a\big]\Big]=\mathbb{E}_a\Big[a^H\mathbb{E}[B]a\Big]=\mathbb{E}_a\Big[a^H\mathbb{E}[bb^H]a\Big]$$
However, in general (e.g. dependent variables, or non-linear $g$), this is not true and we need to use the law of total expectation with the conditional expectation (and corresponding conditional distribution), instead of just replacing the expectation operator when convenient. Note that in my (short) answer above, I needed not check for linearity, as the proper application of law of total expectation and independence of $a,b$ yields the correct answer.