The martingale $M_t$ is not uniformly integrable.
If it were, then by standard martingale facts, it would converge a.s. and in $L^1$ to some $M_\infty$. (An $L^1$-bounded martingale always converges a.s., and if it's uniformly integrable, then the convergence is also $L^1$ by Vitali.) In particular we would have $E[M_\infty] = \lim E[M_t] = 1$.
However, this martingale converges a.s. to zero, and that's a contradiction.
One way to see $M_t \to 0$ is to note, after simplifying, that
$$M_t = \exp\left(\theta \sigma W_t - \frac{\theta^2 \sigma^2 t}{2}\right).$$
By the strong law of large numbers, $W_t / t \to 0$ a.s., and hence
$$\frac{1}{t} \left(\theta \sigma W_t - \frac{\theta^2 \sigma^2 t}{2}\right) \to -\frac{\theta^2 \sigma^2}{2} < 0$$
which implies
$$\theta \sigma W_t - \frac{\theta^2 \sigma^2 t}{2} \to -\infty.$$
It might be tempting to reason as follows: Since $(X_t)_{t \geq 0}$ and $(W_t)_{t \geq 0}$ are Brownian motions, we know that $W_t$ and $X_t$ are Gaussian. On the other hand, it is possible to show (see @Addsup's comment) that $\mathbb{E}(X_t W_t) = 0$, and consequently $X_t$ and $W_t$ are uncorrelated. Since uncorrelated Gaussian random variables are independent, it follows that $X_t$ and $W_t$ are independent.
The reasoning is wrong. Why? In order to conclude that $W_t$ and $X_t$ are independent, we need to know that the random vector $(X_t,W_t)$ is Gaussian; it is not enough to know that (the marginals) $X_t$ and $W_t$ are Gaussian.
In fact, $W_t$ and $X_t$ are not independent. By Itô's formula, we have $$W_t^2 = 2 \int_0^t W_s \,dW_s + t.$$ As $\mathbb{E}(X_t)=0$ we thus find
\begin{align*} \mathbb{E}(W_t^2 X_t)& =2 \mathbb{E} \left( X_t \int_0^t W_s \, dW_s \right) \\ &=2 \mathbb{E} \left(\left[ \int_0^t \text{sgn}(W_s) \, dW_s \right] \left[ \int_0^t W_s \, dW_s \right] \right) \end{align*}
Applying Itô's isometry we obtain that
$$\mathbb{E}(W_t^2 X_t) =2 \mathbb{E} \left( \int_0^t W_s \, \text{sgn}(W_s) \, ds \right) =2\int_0^t \mathbb{E}(|W_s|) \, ds.$$
The integral on the right-hand side is strictly positive (in fact, it can be calculated explicitly, using the fact that $W_s \sim N(0,s)$ entails $\mathbb{E}(|W_s|) = \sqrt{(2s)/\pi}$). As $\mathbb{E}(X_t)=0$ this shows that $$\mathbb{E}(W_t^2 X_t) \neq \mathbb{E}(W_t^2) \mathbb{E}(X_t),$$ and therefore $W_t$ and $X_t$ are not independent.
Best Answer
Brownian motion, Solution I
Since $W_t \sim N(0,t)$, we have
$$\mathbb{E}(|W_t| 1_{\{|W_t|>K\}}) = \frac{2}{\sqrt{2\pi t}} \int_K^{\infty} x \exp \left( -\frac{x^2}{2t} \right) \, dx = \sqrt{\frac{2}{\pi}} \sqrt{t} \exp \left(-\frac{K^2}{2t} \right)$$
and therefore
$$\sup_{t \geq 0} \mathbb{E}(|W_t| 1_{\{|W_t|>K\}})=\infty.$$
Brownian motion, Solution II
Fix $K \geq 1$. Then, using that $W_t \stackrel{d}{=} \sqrt{t} W_1$,
$$\mathbb{E}(|W_t| 1_{\{|W_t|>K\}}) \geq \mathbb{P}(|W_t|>K) = \mathbb{P}\left(|W_1| > \frac{K}{\sqrt{t}} \right) \stackrel{t \to \infty}{\to} 1.$$
Consequently,
$$\lim_{K \to \infty} \sup_{t \geq 0} \mathbb{E}(|W_t| 1_{\{|W_t|>K\}})\geq 1.$$
(Non)Uniform integrability of $M_t := W_t^2-t$
For $K \geq 1$, we have by the triangle inequality
$$\mathbb{E}(|M_t| 1_{\{|M_t|>K\}}) \geq \mathbb{P}(|M_t|>K) \geq \mathbb{P}(|W_t|^2 > K-t).$$
Using again the scaling property, we find
$$\mathbb{E}(|M_t| 1_{\{|W_t|>K\}}) \geq \mathbb{P} \left( |W_1|^2 > \frac{K-t}{t} \right). \tag{1}$$
Choosing $t = K/2$, we get
$$\begin{align*}\lim_{K \to \infty} \sup_{t \geq 0} \mathbb{E}(|M_t| 1_{\{|M_t|>K\}}) &\geq \liminf_{K \to \infty} \mathbb{E}(|M_{K/2}| 1_{\{|M_{K/2}|>K\}}) \\ &\stackrel{(1)}{\geq} \mathbb{P}(|W_1|^2 > 1) >0. \end{align*}$$
This shows that $(M_t)_{t \geq 0}$ is not uniformly integrable.