Overwhelming computer evidence suggests the following, but I would like to see a proof. Let $p\geq 5$ be a prime number. Then, for $0\leq n \leq p-4$, $$\binom{p-2}{n+2}^2\binom{p+n}{n+2}^2-\frac{p^2(n+3)^2}{(n+2)^2(n+1)^2}= \frac{p^3a}{b}$$ where $a$ and $b$ are integers and $p$ does not divide $b$.
Update: This question arose from my reading of a paper where this fact was used without proof in order to derive congruences module $p^3$ of certain combinations of binomial coefficients. A few days after posting the question I was able to find the solution, so I answered my own question.
First note that combining the two fractions $$\frac{(p-2)^2(p-3)^2\cdots (p-n-3)^2(p+n)^2(p+n-1)^2\cdots (p-1)^2}{((n+2)!)^4}-\frac{p^2(n+3)^2}{(n+2)^2(n+1)^2},$$ all factors in the denominator are smaller than $p$. Now think of the numerator of the first fraction as a polynomial in the indeterminate $p$. We can factor out $p^2$ and get the polynomial $$(p-2)^2(p-3)^2\cdots (p-n-3)^2(p+n)^2(p+n-1)^2\cdots (p+1)^2(p-1)^2.$$ Setting $p=0$, the constant term of this polynomial is $((n+3)!)^2 (n!)^2$. Using $((n+2)!)^4$ as the denominator for the second fraction, the numerator becomes $p^2((n+3)!)^2(n!)^2$, so when combining the two fractions, the coefficients of $p^2$ will cancel and we are left with a polynomial that is a factor of $p^3$. Going back to thinking of $p$ as a prime number, it cannot divide the denominator, so the result follows.