Step 1: Name the roots of the first polynomial.
Let the roots of $f(x) = x^2 + mx + p$ be $\alpha$ and $\beta$. We are told they are distinct, negative, and rational. By Vieta, $\alpha + \beta = -m$ and $\alpha\beta = p$.
Step 2: Rewrite the coefficients of $g(x)$ using $\alpha, \beta$.
The middle coefficient of $g$ is $m^2 - 2p$. Since $m^2 = (\alpha + \beta)^2$ and $2p = 2\alpha\beta$, we get $m^2 - 2p = (\alpha + \beta)^2 - 2\alpha\beta = \alpha^2 + \beta^2$. Also the constant of $g$ is $p^2 = (\alpha\beta)^2$.
Step 3: Read off the roots of $g$.
So $g(x) = x^2 - (\alpha^2 + \beta^2)x + (\alpha\beta)^2$. The sum of its roots is $\alpha^2 + \beta^2$ and the product is $\alpha^2 \beta^2$. This factors as $g(x) = (x - \alpha^2)(x - \beta^2)$, so the roots are $\alpha^2$ and $\beta^2$.
Step 4: Check the sign of the roots.
Since $\alpha$ and $\beta$ are real and nonzero, their squares $\alpha^2$ and $\beta^2$ are strictly positive.
Step 5: Check they are distinct.
Because $\alpha$ and $\beta$ are both negative and unequal, their squares stay unequal (two different negatives cannot share the same square). So $\alpha^2 \ne \beta^2$.
Step 6: Check rationality.
The squares of rational numbers are rational, so $\alpha^2$ and $\beta^2$ are rational. Hence $g(x)$ has distinct positive rational roots.
\[ \boxed{\text{positive rational roots}} \]