Given that two samples $\{X_i\}$ and $\{Y_j\}$ are also independent (you implied this condition in your title and attempt but didn't make it explicit in the problem statement),
\begin{align*}
& \sqrt{n_1}(\hat{p}_1 - p_1) \to_d Z_1 \sim N(0, p_1(1 - p_1)), \text{ as } n_1 \to \infty, \tag{1}\label{1} \\
& \sqrt{n_2}(\hat{p}_2 - p_2) \to_d Z_2 \sim N(0, p_2(1 - p_2)), \text{ as } n_2 \to \infty, \tag{2}\label{2}
\end{align*}
as well as $\frac{n_i}{n} \to \lambda_i \in (0, 1)$ as $n \to \infty$, intuitively, it follows that
\begin{align*}
& \sqrt{n}[(\hat{p}_1 - \hat{p}_2) - (p_1 - p_2)] \\
=& \sqrt{n}(\hat{p}_1 - p_1) - \sqrt{n}(\hat{p}_2 - p_2) \\
=& \sqrt{\frac{n}{n_1}}\sqrt{n_1}(\hat{p}_1 - p_1) - \sqrt{\frac{n}{n_2}}\sqrt{n_2}(\hat{p}_2 - p_2) \\
\to_d & \sqrt{\lambda_1^{-1}}Z_1 - \sqrt{\lambda_2^{-1}}Z_2 \sim N(0, \lambda_1^{-1}p_1(1 - p_1) + \lambda_2^{-1}p_2(1 - p_2)) \tag{3}\label{3}
\end{align*}
as $\color{red}{n \to \infty}$. Hence the conclusion holds. Clearly, the last "$\to$" step needs additional justifications.
First, in order to apply $\eqref{1}$ and $\eqref{2}$ as $\color{red}{n \to \infty}$ , we need to ensure that $n_i \to \infty$ as $n \to \infty$, which follows by writing $n_i = \frac{n_i}{n} \times n$ and the condition $n_i/n \to \lambda_i \color{red}{>} 0$.
Then, the Slutsky's theorem guarantees that for $i = 1, 2$:
\begin{align*}
\sqrt{\frac{n}{n_i}}\sqrt{n_i}(\hat{p}_1 - p_1) \to_d \sqrt{\lambda_i^{-1}}Z_i \tag{4}\label{4}
\end{align*}
as $n \to \infty$.
Now if $Z_1$ and $Z_2$ are independent, then the "$\sim$" relation in $\eqref{3}$ holds, and the proof is complete. However, it seems quite difficult to give it a rigorous proof. This motivates me to consider exploiting the independence between $\{X_i\}$ and $\{Y_j\}$ from a different angle.
By Lévy's continuity theorem, to show $\eqref{3}$, it is equivalent to show the characteristic function of the random variable $\sqrt{n}(\hat{p}_1 - p_1) - \sqrt{n}(\hat{p}_2 - p_2)$ converges point-wisely $e^{-\frac{1}{2}\left(\frac{p_1(1 - p_1)}{\lambda_1} + \frac{p_2(1 - p_2)}{\lambda_2}\right)t^2}$, which is the characteristic function of a $N(0, \lambda_1^{-1}p_1(1 - p_1) + \lambda_2^{-1}p_2(1 - p_2))$ random variable. To this end, we have
\begin{align*}
& E\left[e^{it\sqrt{n}((\hat{p}_1 - p_1) - (\hat{p}_2 - p_2))}\right] \\
=& E\left[e^{it\sqrt{n}(\hat{p}_1 - p_1)}\right]E\left[e^{i(-t)\sqrt{n}(\hat{p}_2 - p_2)}\right] \tag{5.1}\label{5.1} \\
=& E\left[e^{it\sqrt{\frac{n}{n_1}}\sqrt{n_1}(\hat{p}_1 - p_1)}\right]E\left[e^{i(-t)\sqrt{\frac{n}{n_2}}\sqrt{n_2}(\hat{p}_2 - p_2)}\right] \\
\to & e^{-\frac{1}{2}\lambda_1^{-1}p_1(1 - p_1)t^2}e^{-\frac{1}{2}\lambda_2^{-1}p_2(1 - p_2)(-t)^2} \tag{5.2}\label{5.2} \\
=& e^{-\frac{1}{2}\left(\frac{p_1(1 - p_1)}{\lambda_1} + \frac{p_2(1 - p_2)}{\lambda_2}\right)t^2}
\end{align*}
as $n \to \infty$. This completes the proof. In $\eqref{5.1}$, we used the condition that $\hat{p}_1$ and $\hat{p}_2$ are independent (which is a consequence of $\{X_i\}$ and $\{Y_j\}$ are independent). In $\eqref{5.2}$, we used the established result $\eqref{4}$ and Lévy's continuity theorem (the other direction).
