Why are rough stochastic volatility models not Markovian? Using the Mandelbrot-Van Ness representation of fractional Brownian motion in terms of Wiener integrals, increments of the realized variance's logarithm $v_t = \sigma_t^{2}$, under the physical measure $\mathcal{P}$, are expressed as
\begin{equation}
\begin{aligned}
\log v_{u}-\log v_{t} &=2 \nu \left(W_{u}^{H}-W_{t}^{H}\right) \\
&=2 \nu C_{H}\left(\int_{-\infty}^{u}|u-s|^{H-\frac{1}{2}} d W_{s}^{\mathbb{P}}-\int_{-\infty}^{t}|t-s|^{H-\frac{1}{2}} d W_{s}^{\mathbb{P}}\right) \\
&=2 \nu C_{H}\left(\int_{t}^{u}|u-s|^{H-\frac{1}{2}} d W_{s}^{\mathbb{P}}+\int_{-\infty}^{t}\left[|u-s|^{H-\frac{1}{2}}-|t-s|^{H-\frac{1}{2}}\right] d W_{s}^{\mathbb{P}}\right) \\
&=: 2 \nu C_{H}\left[M_{t}(u)+Z_{t}(u)\right]
\end{aligned}
\end{equation}
With $H$, our Hurst parameter that determines the roughness of the fractional Brownian motion. In this expression, the left integral $M_{t}(u)$ is independent of the filtration $\mathcal{F}_{t}$ and the right integral $Z_{t}(u)$ is $\mathcal{F}_{t}$-measurable. Note that $\tilde{W}^{P}$ is defined as:
\begin{equation}
\tilde{W}^{P}:=\sqrt{2 H} \int_{t}^{u} \frac{d W_{s}^{\mathbb{P}}}{(u-s)^{\frac{1}{2} -H}}
\end{equation}
On this step I understand that  why we separated $\tilde{W}^{P}$ from the rest thanks to the definition of the weak stationary fractional Brownian motion of Lévy:
\begin{equation}
W_{t}^{H} = L_{H} \int_{0}^{t}(t-s)^{H-\frac{1}{2}} d W_{s}
\end{equation}
With $L_{H}$ being equal to :
\begin{equation*}
L_{H} = \sqrt{2 H}
\end{equation*}
This can be easily solved by computing the variance of the process such, knowing that the covariance of such process under the Gaussian distribution  is equal to:
\begin{equation*}
E(W^{H}_{u}W^{H}_{t}) = (u - t)^{2H}
\end{equation*}
To continue it is said that $\tilde{W}^{P}$ has the same properties as $M_{t}(u)$, only with variance $(u − t)^{2H}$ instead of $\frac{(u − t)^{2H}}{2H}$ . With $\eta:=\frac{2\nu C_{H}}{\sqrt{2H}}$ we have $2\nu M_{t}(u) C_{H}= \eta \tilde{W}^{P}$ and so:
\begin{equation}
\mathbb{E}^{\mathbb{P}}\left[v_{u} \mid \mathcal{F}_{t}\right]=v_{t} \exp \left\{2 \nu C_{H} Z_{t}(u)+\frac{1}{2} \eta^{2} \mathbb{E}\left|\tilde{W}_{t}^{\mathbb{P}}(u)\right|^{2}\right\}
\end{equation}
However, I do not clearly understand this passage. Is it because $Z_{t}(u)$ depends only on historical values, which makes it non-Markovian that we do not treat it as a random variable here? Could someone please give me a more or less rigorous explanation about what happened here. After that, the last step is straightforward to derive as:
\begin{equation}
\begin{aligned}
v_{u} &=v_{t} \exp \left\{\eta \tilde{W}_{t}^{\mathbb{P}}(u)+2 \nu C_{H} Z_{t}(u)\right\} \\
\Rightarrow  v_{u} &=\mathbb{E}^{\mathbb{P}}\left[v_{u} \mid \mathcal{F}_{t}\right] \mathcal{E}\left(\eta \tilde{W}_{t}^{\mathbb{P}}(u)\right)
\end{aligned}
\end{equation}
With $\mathcal{E}$ being the Wick stochastic integral such :
\begin{equation}
\mathcal{E}(\Psi)=\exp \left(\Psi-\frac{1}{2} \mathbb{E}\left[|\Psi|^{2}\right]\right)
\end{equation}
Lastly I also don't understand why in this rough volatility model we have $$\mathbb{E}^{\mathbb{P}}\left[v_{u} \mid \mathcal{F}_{t}\right] \neq \mathbb{E}^{\mathbb{P}}\left[v_{u} \mid v_{t}\right]$$
For more information about the process in details you can read the page 8 and 9 of the following paper: Pricing under rough volatility.
 A: Computing $\mathbb{E}(v_u\,| \,\mathcal{F}_t)$
From your first sequence of formal manipulations and the definition of $\tilde{W}_t^{\mathbb{P}}(u)$ it follows that for $u > t$,
$$v_u = v_t \exp \left( \eta \tilde{W}_t^{\mathbb{P}}(u) + 2 \nu C_H Z_t(u)  \right) = v_t \exp \left( \eta \tilde{W}_t^{\mathbb{P}}(u) \right) \exp\left( 2 \nu C_H Z_t(u)  \right)$$
where, in particular, $Z_t(u) = \int_{-\infty}^{t}\left[|u-s|^{H-\frac{1}{2}}-|t-s|^{H-\frac{1}{2}}\right] d W_{s}^{\mathbb{P}}$. The process $Z_t$ is the Wiener integral of a (deterministic) power law kernel. The authors implicitly define $\mathcal{F}_t$ to be the filtration generated by the Brownian motion on the line, which implies that this Wiener integral is measurable with respect to the Brownian filtration.
Now, seeing that $\tilde{W}_{t}^{\mathbb{P}}(u)$ depends only on the evolution of Brownian motion on $[t,u]$, this process is independent of $\mathcal{F}_t$. Using properties of conditional expectation, it follows that:
$$\begin{align*}
\mathbb{E}(v_u\,| \,\mathcal{F}_t) &= v_t \mathbb{E}\left(\exp \left( \eta \tilde{W}_t^{\mathbb{P}}(u) \right) \exp\left( 2 \nu C_H Z_t(u)  \right)\,| \,\mathcal{F}_t\right) \\
&= v_t \exp\left( 2 \nu C_H Z_t(u)  \right) \mathbb{E}\left(\exp \left( \eta \tilde{W}_t^{\mathbb{P}}(u) \right) \,| \,\mathcal{F}_t\right) \\
&= v_t \exp\left( 2 \nu C_H Z_t(u)  \right) \mathbb{E}\left(\exp \left( \eta \tilde{W}_t^{\mathbb{P}}(u) \right) \right) \\
&= v_t \exp\left( 2 \nu C_H Z_t(u)  \right) \exp \left( \frac{1}{2} \eta^2 \mathbb{E}| \tilde{W}_t^{\mathbb{P}}(u) |^2 \right) \\
&= v_t \exp\left( 2 \nu C_H Z_t(u)  + \frac{1}{2} \eta^2 \mathbb{E}| \tilde{W}_t^{\mathbb{P}}(u) |^2 \right)
\end{align*}$$
$v_t$ is not Markovian
Intuitively, the reason for $\mathbb{E}(v_u\,| \,\mathcal{F}_t) \neq \mathbb{E}(v_u\,| \,v_t)$ is that $Z_t(u)$ depends on the entire history (back to the Big Bang, as one of the authors of that paper likes to say) of Brownian motion. If we tried to retrace the steps in the previous computation, we would need to compute the term
$$\mathbb{E}\left( \exp(2 \nu C_H Z_t(u)) \, | \, v_t\right)$$
In general, there is no hope that $Z_t(u)$ is $\sigma(v_t)$-measurable, implying that $$\mathbb{E}\left( \exp(2 \nu C_H Z_t(u)) \, | \, v_t\right) \neq \exp(2 \nu C_H Z_t(u))$$
