Since $(Y_t)_{t \geq 0}$ is a bounded submartingale, there exists $Y_{\infty} \in L^1$ such that $Y_t \to Y_{\infty}$ in $L^1$ as $t \to \infty$. Moreover, by the monotone convergence theorem,
$$\mathbb{E}\left( \sup_{t \geq 0} A_t \right) = \sup_{t \geq 0} \mathbb{E}(A_t),$$
i.e. $A_t \to A_{\infty} := \sup_{t \geq 0} A_t$ in $L^1$. The decomposition
$$Y_t = Y_0 + M_t+A_t$$
yields $M_t \to M_{\infty}$ in $L^1$ for some random variable $M_{\infty}$ and $(M_t)_{t \in (0,\infty]}$ is a martingale; in particular uniformly integrable. Now the following theorem applies:
Theorem Let $Y_t = M_t+A_t$ where $(M_t)_{t \geq 0}$ is a uniformly integrable martingale and $(A_t)_t$ an increasing predictable process with $A_0=0$ a.s. If $(Y_t)_t$ is bounded, then $\mathbb{E}(A_{\infty}^p)<\infty$ for any $p>0$.
Proof: Fix $C>0$ such that $|Y_t| \leq C$ for all $t \geq 0$. For $\lambda>0$ we define a stopping time $$T:= \inf\{t \geq 0; A_t \geq \lambda\}.$$ Then
$$\begin{align*} \mathbb{P}(A_{\infty} \geq \lambda+4C) &= \mathbb{P}(A_{\infty} \geq \lambda+C, T<\infty) \leq \mathbb{P}(A_{\infty}-A_T \geq 4C, T<\infty) \\ &\leq \frac{1}{4C} \mathbb{E}((A_{\infty}-A_T) \cdot 1_{\{T<\infty\}}). \tag{1}\end{align*}$$
Now let $$T_n := \inf\{t \geq 0; A_t \geq \lambda-1/n\}.$$
Then $T_n \uparrow T$ and $\mathcal{F}_{T_n} \subseteq \mathcal{F}_T$. For $j \geq k$, we have $\{T_k <N\} \in \mathcal{F}_{T_k} \subseteq \mathcal{F}_{t_j}$ and therefore, by the tower property,
$$\begin{align*} \mathbb{E}((A_{\infty}-A_{T_j}) \cdot 1_{\{T_k<N\}}) &= \mathbb{E}\bigg[ \mathbb{E}(A_{\infty}-A_T \mid \mathcal{F}_{T_k}) \cdot 1_{\{T_k<N\}} \bigg] \\ &= \mathbb{E}\bigg[ \mathbb{E}(Y_{\infty}-Y_T \mid \mathcal{F}_{T_k}) \cdot 1_{\{T_j<N\}} \bigg] \\ &\leq 2C \mathbb{P}(T_k<N). \end{align*}$$
Here we used in the second step that $-M_t = A_t-Y_t$ is a uniformly integrable martingale. Applying Fatou's lemma yields for $j \to \infty$
$$ \mathbb{E}((A_{\infty}-A_T) \cdot 1_{\{T_k<N\}}) \leq 2C \mathbb{P}(T_k<N).$$
Applying again Fatou's lemma (for $k \to \infty$) we see that
$$\mathbb{E}((A_{\infty}-A_T) \cdot 1_{\{T<N\}}) \leq 2 C \mathbb{P}(T \leq N).$$
Finally, by the monotone convergence theorem,
$$\mathbb{E}((A_{\infty}-A_T) \cdot 1_{\{T<\infty\}}) \leq 2 C \mathbb{P}(T < \infty). \tag{2}$$
Combining $(1)$ and $(2)$ we obtain
$$\mathbb{P}(A_{\infty} \geq \lambda+4C) \leq \frac{1}{2} \mathbb{P}(T<\infty) = \frac{1}{2} \mathbb{P}(A_{\infty} \geq \lambda).$$
In particular, for $\lambda = 4kC$, $k \in \mathbb{N}_0$,
$$\mathbb{P}(A_{\infty} \geq (k+1) 4C) \leq \frac{1}{2} \mathbb{P}(A_{\infty} \geq 4kC).$$
Iterating this inequality yields
$$\mathbb{P}(A_{\infty} \geq 4kC) \leq \frac{1}{2^{k-1}}.$$
Now the claim follows from the fact that
$$\mathbb{E}(A_\infty^p) = p \int_{(0,\infty)} x^{p-1} \mathbb{P}(A_{\infty} \geq x) \, dx.$$
Remark The proof is adapted from Proposition 16.32 in Stochastic Processes by Richard F. Bass.
(... and I guess that there is an easier (or more elegant) way to prove the claim; this one seems rather like overkill to me.)
