The answer to both questions is "no" - a counterexample (labeled "folklore") appears immediately prior to Question 6.6 in Austin, Tim, Extensions of probability-preserving systems by measurably-varying homogeneous spaces and applications, Fundam. Math. 210, No. 2, 133-206 (2010). ZBL1206.28023..
Here it is, essentially verbatim:
Let $G = \mathbb Z\times \mathbb Z$, let $\mathbb T = \mathbb R/\mathbb Z$ with Haar probability measure $m$, and let $X = \mathbb T^2 \times \mathbb T^2$ and $\mu=$ Haar probability measure on $X$. Let $A:\mathbb T^2\to \mathbb T^2$ be any ergodic toral automorphism, and define an action of $\mathbb Z\times \mathbb Z$ on $X$ by $(n,m)\cdot (\mathbf x, \mathbf y) = (A^n\mathbf x, A^n(\mathbf y+m\mathbf x))$. In other words, $G \curvearrowright (X,\mu)$ is the action generated by $A\times A$ and the map $(\mathbf x,\mathbf y)\mapsto (\mathbf x, \mathbf y+\mathbf x)$. The action generated by $A\times A$ is ergodic (and in fact mixing), since the action generated by $A$ is mixing. The entire action is therefore weak mixing.
Taking $H:= \{(0,m):m\in \mathbb Z\}$, the ergodic components are given by rotation $(0,m)\cdot \mathbf y = \mathbf y + m \mathbf x$ on the vertical $Y_{\mathbf x}:=\{(\mathbf x, \mathbf y): \mathbf y\in \mathbb T^2\}$. Examining the spectrum of these components reveals that they are not (almost all) mutually isomorphic, and no component is weak mixing.