The minimum value of $p$ is $5$.

Write $a$ and $b$ as fractions with least common denominator $w$: $a = \frac{u}{w}$, $b = \frac{v}{w}$. In other words, if $a = \frac{u'}{w'}$, $b = \frac{v'}{w'}$ is another representation with common denominator $w'$, then $w' \ge w$. The irreducible representation $s = \frac{u+v}{w}$ is obtained from $s = \frac{u+v}{w}$ by possible cancelation. Therefore, the prime divisors of $n$ are among the ones of $w$.

We show that $w$ is not divisible by $2$ and $3$, implying that neither is $n$.

The condition $a+b = a^2 + b^2$ gives $u^2 + v^2 = w(u+v)$.

Suppose that $3$ divides $w$. Then $u^2 + v^2$ is a multiple of $3$, and since $x^2 = 0, 1 \pmod{3}$ for each integer $x$, it follows that both $u$ and $v$ are divisible by $3$. However, then $3$ is a common divisor of $u$, $v$, and $w$, which contradicts the minimality of $w$.

Similarly, suppose that $w$ is even. Then $u^2 + v^2$ is even, hence so is $u+v$ ($u^2 + v^2$ has the same parity as $u+v$). Hence $u^2 + v^2$ is divisible by $4$, and since $x^2 = 0, 1 \pmod{4}$ for each integer $x$, both $u$ and $v$ are even. We reach a contradiction with the minimality of $w$ again.

The minimum value of $p$ is $p = 5$. Write $a$ and $b$ as fractions with least common denominator $w$: $a = \frac{u}{w}$, $b = \frac{v}{w}$. In other words, if $a = \frac{u'}{w'}$, $b = \frac{v'}{w'}$ is another representation with common denominator $w'$, then $w' \geq w$. The irreducible representation $s = \frac{m}{n}$ is obtained from $s = \frac{u+v}{w}$ by possible cancellation. Therefore, the prime divisors of $n$ are among the ones of $w$.

We show that $w$ is not divisible by $2$ and $3$, implying that neither is $n$. The condition $a+b = a^2 + b^2$ gives $u^2 + v^2 = w(u+v)$. Suppose that $3$ divides $w$. Then $u^2 + v^2$ is a multiple of $3$, and since $x^2 \equiv 0,1 \pmod{3}$ for each integer $x$, it follows that both $u$ and $v$ are divisible by $3$. However, then $3$ is a common divisor of $u$, $v$, and $w$, which contradicts the minimality of $w$. Similarly, suppose that $w$ is even. Then $u^2 + v^2$ is even, hence so is $u+v$ ($u^2 + v^2$ has the same parity as $u+v$). Hence $u^2 + v^2$ is divisible by $4$, and since $x^2 \equiv 0,1 \pmod{4}$ for each integer $x$, both $u$ and $v$ are even. We reach a contradiction with the minimality of $w$ again.

By the above, each prime divisor of $n$ is at least $5$. For an example with $p = 5$, let $a = \frac{2}{5}$, $b = \frac{6}{5}$.
Then $a+b = \frac{8}{5}$, $a^2 + b^2 = \frac{4}{25} + \frac{36}{25} = \frac{40}{25} = \frac{8}{5}$. So $a+b = a^2 + b^2$ holds, the common value $s$ is not an integer, and its representation $s = \frac{8}{5}$ is irreducible with $p = n = 5$.