Here is one alternative set of conditions. Essentially, I move the $L^p$-uniform integrability requirement of $X_k$ to the predictable component $A_k$ given by Doob's decomposition theorem.

If $X_k$ is submartingale then, by Doob's decompositon theorem, we have $X_k = M_k + A_k$ where $M_k$ is martingale and $A_k$ is a predictable nondecreasing process.

If we assume $\sup_k E|M_k|^p < \infty$ then as you mentioned we would have $M_k \rightarrow M_{\infty}$ in $L^p$ for some limiting variable $M_{\infty}$. So now we need to make an assumption to ensure that $A_k$ converges in $L^p$. Establishing almost sure convergence of $A_k$ isn't difficult (e.g. exploit nondecreasingness). To strengthen the almost sure convergence of $A_k$ to $L^p$, we assume $A_k$ is uniformly integrable in $L^p$. This also implies that $\sup_k E|X_k|^p\lessapprox \sup_k E|M_k + A_k|^p \lessapprox \sup_k E|M_k|^p + \sup_kE|A_k| < \infty$, so as mentioned in your reference, we have $X_k \rightarrow X_{\infty}$ almost surely. But, this implies $A_k = X_k - M_k \rightarrow X_{\infty} - M_{\infty}$ almost surely, so that this condition also suffices to get almost sure convergence of $A_k$ (which we needed).

So one set of conditions, would be

 1. $\sup_k E|M_k|^p < \infty$
 2. $A_k$ is uniformly integrable in $L^p$


Condition 1 is analogous to what is needed when $X_k$ is martingale. Condition 2 requires the predictable part of $X_k$, $A_k$, to be uniformly integrable. As argued above, these conditions imply that $\sup_k E|X_k|^p < \infty$, but this condition itself is not sufficient. Also, noting that condition 1 is equivalent to uniform integrability (U.I.) of $M_k$ and sums of U.I. sequences are U.I., these conditions actually imply that $X_k$ is U.I.

I imagine in practice that the predictable component $A_k$ is simpler than $X_k$ or $M_k$, so verifying its uniform integrability may be significantly easier in some cases than establishing the uniform integrability of $X_k$.