First of all, I am assuming that what you denote $v({\mathfrak {p}})$ should be a function (which I will denote $v_{\mathfrak {p}}$) from the ideals of $R$ to $\mathbb{Z}$ (in fact this function can be extended to fractional ideals). This function is defined as $$I=\mathfrak {p}^{v_{\mathfrak {p}}(I)} J$$
for $J$ an ideal prime to ${\mathfrak {p}}$ (that it is well defined is consequence of some results on Dedekind domains).
Another possibility is first to show that the localization $R_{\mathfrak {p}}$ of $R$ at ${\mathfrak {p}}$ is a discrete valuation ring, which implies that the only ideals are the powers of ${\mathfrak {p}}R_{\mathfrak {p}}$, and define the corresponding power of $I R_{\mathfrak {p}}$ as the valuation of $I$ at ${\mathfrak {p}}$.
The first definition gives an example of discrete valuation defined on the field of fractions of a Dedekind domain, by defining the valuation of an element $x$ of $R$ as $v_{\mathfrak {p}}(xR)$, and extending to the fraction field $K$ by $v_{\mathfrak {p}}(x/y)=v_{\mathfrak {p}}(x)-v_{\mathfrak {p}}(y)$.
But most fields are not of these type, and they could have discrete valuations constructed in another way. For example, if $K=k(t)$, $t$ an indeterminate and $k$ a field, then minus the degree gives a discrete valuation: if $g(t)=p(t)/q(t)$, with $p(t)$ and $q(t)\in k[t]$, then $$v(g(t)):=\deg(q(t))-\deg(p(t)),$$
where $\deg$ denotes the degree on $t$, is a valuation. It does not come from a prime ideal in $k[t]$, which in turn is not a Dedekind domain unless $k$ is finite.
Be careful: $A_\mathfrak{p}$ need not be an integral domain, so you don't want to use the phrase field of fractions there. But we can instead use $k(\mathfrak p):=\operatorname{Frac}(A/\mathfrak{p})$.
On one hand, as you say, if $A\to V$ is a map to a valuation ring then you can restrict the valuation $V$ naturally has to a valuation which is $\le 1$ on $A$.
On the other hand, if you have a valuation $v$ which is $\le1$ on $A$, then it induces a valuation (let's also denote it $v$) on $k(\mathfrak p)$ for $\mathfrak p=\ker(v)$. If you let $V$ be the valuation ring of $v$ inside $k(\mathfrak p)$, i.e. $V=\{x\in k(\mathfrak p)\mid v(x)\le1\}$, then you have $A/\mathfrak p\subseteq V$, so you get a natural map $A\to V$ and this is the map to a valuation ring we want.
We need to know this respects equivalence in both directions. The following will be useful for reasoning this out:
Lemma: A map $V\to W$ of valuation rings is faithfully flat if and only if it is injective and local.
One one hand, it is simple that if $v$ and $w$ are equivalent valuations on $A$ then $\ker(v)=\ker(w)=:\mathfrak p$ and they have the same valuation ring inside $k(\mathfrak p)$, so the map $A\to V$ is unambiguous.
Conversely, if we have $A\to V\to W$ with $V\to W$ faithfully flat, then using our lemma above one can see that $W':=\operatorname{Frac}(V)\cap W$ is a valuation ring of $\operatorname{Frac}(V)$ containing $V$ and also with $\mathfrak m_V\subseteq\mathfrak m_{W'}$, and from this one can deduce that $\operatorname{Frac}(V)\cap W=V$. From this you can deduce that the valuations $v$ and $w$ induced on $A$ by $A\to V$ and $A\to W$ are equivalent.
Now we want to know our two maps are inverse to each other: as you've pretty much noted, if $v$ is a valuation on $A$ and you take the induced map to a valuation ring $A\to V$ we described, then the valuation you get back from this is the original valuation $v$.
Conversely, let's say we started with a map $A\to V$ and we get the induced valuation on $A$. For $\mathfrak p=\ker(v)$ we have the valuation ring $V'$ of $k(\mathfrak p)$ coming from $v$, and we want to show that our maps $A\to V$ and $A\to V'$ are equivalent. We have that $A/\mathfrak p\hookrightarrow V$ is injective, so we have an induced map of fields of fractions $k(\mathfrak p)\hookrightarrow\operatorname{Frac}(V)$ under which one can see that $V'\hookrightarrow V$. Also because $v$ by definition came from $V$, one can see that $\mathfrak m_{V'}\hookrightarrow\mathfrak m_V$. Thus using our lemma above we see that $V'\to V$ is faithfully flat, which is what we wanted.
Best Answer
The important point is that it's not assumed just that $|\cdot|$ is a valuation on $A/\mathfrak p$ but rather it's a valuation on $k(\mathfrak p)=\operatorname{Frac}(A/\mathfrak p)$. Valuations on fields automatically have $\operatorname{supp}(|\cdot|)=\{0\}$ (because the support is a prime ideal).
To see why the claim as originally stated can't be true, it's important to note that if $A$ is an integral domain you could take $\mathfrak p=(0)$ and then the claim would be that every valuation on an integral domain has zero support, for which you have simple counterexamples on, say, $A=\mathbb Z$.