Let $(X^n),(Y^n)$ be a random variables with $E[X^n]\leq x \, \forall n $ for some fixed $x\in \mathbb{R}$ and $X^n\geq Y^n\, \forall n$.
Also assume that $Y^n\rightarrow Y$ in $L^1(P)$ as $n\rightarrow \infty$ for some random variable $Y$.
Does that imply that $\big((X^n)^-\big)_{n\in \mathbb{N}}$ is uniformly integrable?
Yes, it does. To prove so, let $\varepsilon > 0$. We must show that there is $\delta > 0$ so that if $\Omega$ is our sample space and $A \subset \Omega$ is measurable with $P(A) < \delta$, then $\int_A (X^n)^- dP< \varepsilon$. For this, we begin by noticing that $(X^n)^- \leq (Y^n)^- \leq |Y^n|$. (This can be proved by casework.) Thus, it suffices to prove that the family $\{ Y^n \}$ is uniformly integrable.
A proof that the family $Y^n$ is uniformly integrable is as follows. (There may be a more elegant solution, but this is what seemed most immediate.) Fix $\varepsilon > 0$. For any measurable $A$ we have $$ \int_{A} |Y^n|\, dP \leq \int_{A} |Y^n - Y|\, dP + \int_{A} |Y| \,dP. \, \, \, (\star) $$ We can make the integral of $|Y|$ less than $\frac{\varepsilon}{2}$ by taking $P(A)$ sufficiently small - say, less than $\delta_1$. This uses absolute continuity of the Lebesgue integral (see this stack exchange answer: Absolute continuity of the Lebesgue integral).
Meanwhile, we have $$ \int_{A} |Y^n - Y|\, dP \leq \int_{\Omega} |Y^n - Y|\, dP. $$ Since $Y^n \to Y$ in $L^1$, there is $N$ so that if $n \geq N$, then $\int_{\Omega} |Y^n - Y|\, dP < \frac{\varepsilon}{2}$. It follows by $(\star)$ that if $P(A) < \delta_1$, then $\int_A |Y^n| \, dP < \varepsilon$ for all $n \geq N$.
To deal with the case that $n < N$, let $S = |Y_1| + \dots |Y_N|$. Then for each $n < N$, we have $$ \int_{A} |Y^n|\, dP + \int_A |Y| \, dP \leq \int_{A} S + \int_{A} |Y|. $$ By the absolute continuity property mentioned above, we can make $\int_{A} S \, dP$ less than $\varepsilon$ by picking an appropriate $\delta_2$. Thus, for $A$ with $P(A) < \delta_2$, we’ll have $\int_A |Y^n| \, dP < \varepsilon$ for all $n < N$.
We can then take $\delta$ to be the minimum of $\delta_1$ and $\delta_2$.