The short version of this answer is:
Degree theory (e.g. the study of the first-order theory of the partial order of Turing degrees) yields examples of non-relativizing statements, although these statements are of course highly technical. One useful tool for understanding this situation is the cone theorem ... which also limits the extent to which this behavior can happen.
There are indeed theorems which don't relativize, about which I'll say more below. First, though, let me observe that there is a sense in which all (reasonable) theorems eventually relativize:
(Martin's cone theorem) Suppose $A$ is a "nicely definable" property of Turing degrees. Then either every sufficiently large Turing degree has $A$ or every sufficiently large Turing degree fails $A$; precisely, there is some degree ${\bf d}$ such that for every degree ${\bf c}\ge_T{\bf d}$ we have $A({\bf c})\leftrightarrow A({\bf d})$.
"Nicely definable" here is a bit technical; specifically, we want the set of elements of $A$ (thought of as a subset of $2^\omega$) to be determined, in the sense that one player or the other has a winning strategy in the associated Gale-Stewart game. I'll briefly observe that $(i)$ every Borel set is determined, and so every property of Turing-invariant property which is expressible in the infinitary logic $\mathcal{L}_{\omega_1,\omega}$ either holds or fails on a cone, and $(ii)$ additional set-theoretic assumptions can push determinacy far beyond the Borel sets. Heuristically, one will not run into a non-determined set of Turing degrees without explicitly trying to.
The proof of the cone theorem can be boiled down to a single sentence:
Supposing $A\subseteq 2^\omega$ is Turing-invariant and determined with a winning strategy $\sigma$ for player $1$, given any $r\ge_T\sigma$ consider the play of $\sigma$ against the "silly" strategy for player $2$ which simply plays the bits of $r$ regardless of what player $1$ does.
OK, this isn't really fair: the difficult aspect of the cone theorem isn't the theorem itself but rather the determinacy results it leans on, since it's not a prior clear that determinacy is something we should expect of "nice" sets in the first place. The clearest proof of Borel determinacy I've seen is in Moschovakis' book, but it's still quite involved. Large cardinals improve things substantially - the proof of analytic determinacy from a measurable cardinal is actually much simpler than Borel determinacy from ZFC - but it's never really easy after the first couple levels of the Borel hierarchy.
Now let's talk about the positive role of the cone theorem in finding non-relativization phenomena.
First, here's an interesting application of the cone theorem to get a dubious example of what you're looking for. By the Arslanov Completeness Criterion, ${\bf 0'}$ is not PA over anything over which it is low; however, since the low basis theorem does relativize, by the cone theorem we know that for all sufficiently large degrees ${\bf a}$ there is some ${\bf b}$ such that ${\bf a'}$ is low an PA over ${\bf b}$. This apparently contradicts the fact that the proof of ACC also relativizes; the saving grace is that the required ${\bf b}$ is not in the interval $[{\bf a},{\bf a'}]$ (and indeed we get an interesting jump inversion counterexample here).
Ultimately the above indicates that we've (fine, I've) relativized the original result incorrectly here: when relativizing a degree-theoretic fact to ${\bf a}$, we should restrict attention to the upper cone $\{{\bf c}: {\bf c}\ge_T{\bf a}\}$ (that is, look at what happens if ${\bf a}$ were to be thought of as ${\bf 0}$). This suggests a natural sub-question of your question:
Is there some sentence $\varphi$ in the language of partial order such that $\mathcal{D}\models\varphi$ but $\mathcal{D}_{\ge_T{\bf a}}\models\neg\varphi$ for all sufficiently large degrees ${\bf a}$?
Here $\mathcal{D}$ (resp. $\mathcal{D}_{\ge_T{\bf a}}$) is the partial order of Turing degrees (resp. Turing degrees $\ge_T{\bf a}$) under $\ge_T$. The language of partial order may seem quite limited at first, but it's actually extremely expressive; in particular, contra early expectations it is known that $\mathcal{D}$ has a highly complicated first-order theory and at most countably many automorphisms (and it's conjectured that $\mathcal{D}$ is rigid, a proposed counterexample by Cooper being generally considered to be flawed).
- Note that the question above treats a precise notion of (non-)relativization, which I think is rather neat! And per the expressivity of $\mathcal{D}$, it's not too restrictive either.
The answer to this sub-question turns out to be yes; this is a consequence of Theorem 3.1 of Nerode and Shore (although they don't quite phrase it this way - also, they include a symbol for the Turing jump, but it was later shown that the jump is definable from the partial order structure alone).
The proof gives such a $\varphi$, but we can also show the existence of such a $\varphi$ using their theorem as a blackbox ... via the cone theorem! Suppose no such $\varphi$ existed; then for each $\varphi\in Th(\mathcal{D})$, the set $T_\varphi=\{{\bf a}: \mathcal{D}_{\ge{\bf a}}\models\varphi\}$ contains a cone. But since the intersection of countably many cones in $\mathcal{D}$ is again a cone, we get a cone of ${\bf a}$ in $\bigcap_{\varphi\in Sent} T_\varphi$ - in other words, a cone of degrees whose upper cones are elementarily equivalent to $\mathcal{D}$, contradicting Nerode/Shore.
