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I have following question about Universal Property for Quotient Field that leads me to following contradiction:

The statement is: Let $R$ be an intergal domain and $phi: R to F$ is a ring morphism. Futhermore, let $i: R to Q(R)$ be the canonic inclusion. Then there exist a unique morphism $barphi: Q(R) to F$ with $phi = barphi circ i$.

Then I consider following example: Let $R = mathbbZ, F = mathbbZ/(p)$ for a prime $p$. Let $phi: mathbbZ to mathbbZ/(p)$ be the canonic projection. Obviously, $phi neq 0$. But on the other hand the extension $barphi: mathbbQ= Q(mathbbZ) to mathbbZ/(p)$ is obviously zero map since in $mathbbQ= Q(mathbbZ)$ we have $1 = fracpp$ and therefore $barphi(1)=0$. Where is the error in my reasonings?

The ring homomorphisms $phi:Rto F$ is required to be injective (and $F$ is required to be a field).
See here for example.
More generally, a ring homomorphism $phi:Rto F$ extends to a ring homomorphism $Q(R)to F$ if and only if $phi$ maps every non-zero element of $R$ into a unit in $F$.

Your statement of the universal property is incorrect. It only applies when the homomorphism $phi:Rto F$ is injective.

The theorem requires that the map sends denominators into units, which fails for denominators that are multiples of $p$. We can work around this if we restrict to fractions having denominators coprime to $p$. Then, by a similar proof, the natural ring map of $,Bbb Z,$ into $,Bbb Z/p,$ extends to the subring of $,Bbb Q,$ of fractions writable with denominator coprime to $p$, via $,a/b mapsto ab^-1pmod p.,$ This implies that it is valid to use such fractions when working in $,Bbb Z/p.,$ You can find many examples of such modular fraction arithmetic in my posts here.