$\newcommand{\Cat}{\mathrm{Cat}_\infty} \newcommand{\Spaces}{\mathrm{Spaces}} \newcommand{\map}{\mathrm{map}} \newcommand{\Fun}{\mathrm{Fun}}$ Any coCartesian fibration which is a localization is a universal localization, since coCartesian fibrations are closed under pullbacks, and the fibers are stable under pullbacks as well.
In particular, any localization which inverts all maps, since it has an $\infty$-groupoid as target, is a coCartesian fibration and is therefore a universal localization.
Any (co)reflexive localization is also universal because (co)reflexive localizations are closed under pullbacks.
So we have to look a bit further for examples of non-universal localizations. An easy thing to come up with an example is to observe that $C[W^{-1}]$ typically has more morphisms than $C$, so the pullback $\Delta^1\times_{C[W^{-1}]}C$ is likely to be too small. Let $p:C\to C[W^{-1}]$ denote the localization functor.
Indeed, given a map $f: x\to y$ in $C[W^{-1}]$, objects of $\Delta^1\times_{C[W^{-1}]} C$ are objects of the fibers $p^{-1}(x), p^{-1}(y)$, there are no morphisms in the direction $y\to x$, and morphisms lying over $0\to 1$ are maps from some $x_0\in p^{-1}(x)$ to some $y_0\in p^{-1}(y)$ lifting $f$.
Consider for example the following : $C$ is $a\to b, a\to c, d\to c$ and nothing else. If you invert $a\to c$, you can prove that the resulting category is $[2]= 0\leq 1\leq 2$ with $p: d\mapsto 0, a,c\mapsto 1, b\mapsto 2$, so that there are no maps from the fiber over $0$ to the fiber over $2$. Thus in this example (which I recommend drawing out), $\Delta^1\times_{C[W^{-1}]}C$ has two objects and no maps in either direction, and so it cannot localize onto $\Delta^1$.
EDIT: I had misread the question, and thought the OP already stated the following converse to the coCartesian fibration case: if a coCartesian fibration is a localization, then it has weakly contractible fibers.
This is in fact true - let $p:E\to B$ be a coCartesian fibration. By un/straightening it corresponds to a functor $B\to \Cat, b\mapsto E_b$. Consider the left fibration $|E|_B\to B$ corresponding to the composite $B\to \Cat\xrightarrow{|-|} \Spaces$. Since there is a natural map $C\to |C|$, there is a natural map $E\to|E|_B$ lying over $B$. I claim that 1) $E\to|E|_B$ is a localization and 2) $|E|_B\to B$ is conservative.
With these two facts, we will be done. Indeed, if $E\to B$ is a localization, then by 1) $|E|_B\to B$ is also one, but it is conservative by 2) and so it must be an equivalence. But its fibers are $|E_b|$, so them being points exactly means that the fibers of $E$ are weakly contractible.
Now 2) is easy: $|E|_B\to B$ is a left fibration and left fibrations are all conservative. For 1), simply note that for $C\in \Cat$, restriction along $E\to |E|_B$ induces a map $\map(|E|_B,C)\to \map(E,C)$ which is equivalent to $\map_{/B}(|E|_B, C\times B) \to \map_{/B}(E,C\times B)$, which is in turn equivalent to $\map_{\Fun(B,Cat)}(|E_\bullet|, C) \to \map_{\Fun(B,Cat)}(E_\bullet, C)$ which is exactly the inclusion of components of all natural transformations $E_\bullet\to C$ that pointwise invert all edges in $E_\bullet$. Through unstraightening, this is saying that $|E|_B\simeq E[p^{-1}(B^\simeq)^{-1}]$.
(A reference for 1) is Hinich's Dwyer Kan localization revisited, Prop 2.1.4)
