No.

Indeed, make the substitution $x=u^{-1/2}$. That is, let $G_n(u):=F_n(u^{-1/2})$ and $G(u):=F(u^{-1/2})$, so that $\int_0^1\frac{F_n(x)}{x^3}\,dx=\int_1^\infty G_n$ and $\int_0^1\frac{F(x)}{x^3}\,dx=\int_1^\infty G$. So, we can restate the question as follows:

Let $(G_n)$ be a sequence of functions on $[1,\infty)$ converging uniformly to a function $G$ on $[1,\infty)$ so that $\int_1^\infty G_n$ converges (as $n\to\infty$) and $\int_1^\infty G$ exists. Does it then follow that $\int_1^\infty G_n\to\int_1^\infty G$?

To show that this is not true, let e.g. $$G_n:=\frac1n\,1_{[n,2n]}.$$ Then $G_n\to0=:G$ uniformly but $$\int_1^\infty G_n=1\not\to0=\int_1^\infty G.$$

No.

Indeed, make the substitution $x=u^{-1/2}$. That is, let $G_n(u):=F_n(u^{-1/2})$ and $G(u):=F(u^{-1/2})$, so that $\int_0^1\frac{F_n(x)}{x^3}\,dx=\frac12\,\int_1^\infty G_n$ and $\int_0^1\frac{F(x)}{x^3}\,dx=\frac12\,\int_1^\infty G$. So, we can restate the question as follows:

Let $(G_n)$ be a sequence of functions on $[1,\infty)$ converging uniformly to a function $G$ on $[1,\infty)$ so that $\int_1^\infty G_n$ converges (as $n\to\infty$) and $\int_1^\infty G$ exists. Does it then follow that $\int_1^\infty G_n\to\int_1^\infty G$?

To show that this is not true, let e.g. $$G_n:=\frac1n\,1_{[n,2n]}.$$ Then $G_n\to0=:G$ uniformly but $$\int_1^\infty G_n=1\not\to0=\int_1^\infty G.$$

No.

Indeed, making the substitution $x=u^{-1/2}$, restate the question as follows:

Let $(G_n)$ be a sequence of functions on $[1,\infty)$ converging uniformly to a function $G$ on $[1,\infty)$ so that $\int_1^\infty G_n$ converges (as $n\to\infty$) and $\int_1^\infty G$ exists. Does it then follow that $\int_1^\infty G_n\to\int_1^\infty G$?

To show that this is not true, let e.g. $$G_n:=\frac1n\,1_{[n,2n]}.$$ Then $G_n\to0=:G$ uniformly but $$\int_1^\infty G_n=1\not\to0=\int_1^\infty G.$$

No.

Indeed, make the substitution $x=u^{-1/2}$. That is, let $G_n(u):=F_n(u^{-1/2})$ and $G(u):=F(u^{-1/2})$, so that $\int_0^1\frac{F_n(x)}{x^3}\,dx=\int_1^\infty G_n$ and $\int_0^1\frac{F(x)}{x^3}\,dx=\int_1^\infty G$. So, we can restate the question as follows:

Let $(G_n)$ be a sequence of functions on $[1,\infty)$ converging uniformly to a function $G$ on $[1,\infty)$ so that $\int_1^\infty G_n$ converges (as $n\to\infty$) and $\int_1^\infty G$ exists. Does it then follow that $\int_1^\infty G_n\to\int_1^\infty G$?

To show that this is not true, let e.g. $$G_n:=\frac1n\,1_{[n,2n]}.$$ Then $G_n\to0=:G$ uniformly but $$\int_1^\infty G_n=1\not\to0=\int_1^\infty G.$$