Existence of Lascar invariant type I am looking for a type $q(x)\in S(A)$ that is not contained in any global type $p(x)\in S({\mathcal U})$ that is Lascar invariant over $A$.
By Lascar invariant I mean that $p(x)$ is invariant over every model containing $A$.
 A: Take as a theory $(\mathbb{Q}, \operatorname{cyc})$, the cyclic order on the rationals. That is, $\operatorname{cyc}(x, y, z)$ is a ternary relation given by
$$
(x < y < z) \vee (y < z < x) \vee (z < x < y).
$$
This theory has quantifier elimination, which we will implicitly use in what follows. The picture to keep in mind is that of a circle, where $\operatorname{cyc}(x, y, z)$ holds if we encounter $x, y, z$ in that order when going clockwise.
Take $A$ to be the empty set. There is only one type $q(x) \in S(\emptyset)$, and we claim that this type does not extend to a global Lascar-invariant type. Suppose for a contradiction that it does. Let $p(x) \supseteq q(x)$ be a global Lascar-invariant extension and let $a$ be a realisation of $p$ (in some bigger monster model). We distinguish three cases, and will show that each case is contradictory.

*

*Either $a = 0$ or $a = 1$. Let $M = \{ b \in \mathbb{Q} : b < 0 \}$ then $M$ is a model and $0 \equiv_M 1$. So the formula $x = 0$ is in $p(x)$ if and only if $x = 1$ is in $p(x)$. Since we can never have that $x = 0$ and $x = 1$ are both in $p(x)$, we must have that $x \neq 0$ and $x \neq 1$ are in $p(x)$, so $a$ cannot be $0$ or $1$.

*In case $\operatorname{cyc}(0, a, 1)$ we let $M$ be as in the first case. We note that $(0, 1) \equiv_M (1, 2)$. So since $\operatorname{cyc}(0, x, 1) \in p(x)$ we must have $\operatorname{cyc}(1, x, 2) \in p(x)$, a contradiction.

*In case $\operatorname{cyc}(1, a, 0)$ we consider the model $N = \{b \in \mathbb{Q} : \frac{1}{2} < b < 1 \}$. Then similar to the second case we have $(1, 0) \equiv_N (0, \frac{1}{2})$. So since $\operatorname{cyc}(1, x, 0) \in p(x)$ we must also have $\operatorname{cyc}(0, x, \frac{1}{2}) \in p(x)$, a contradiction.

A: Mark Kamsma's answer gives an example in an NIP theory. As you pointed out in the comments: In an NIP theory, a global type does not fork over $A$ if and only if is Lascar invariant over $A$. So $q(x)\in S(A)$ forks over $A$ if and only if there is no global extension of $q$ which is Lascar invariant over $A$.
There is no such type in a stable theory, since stable theories are NIP, and no type in $S(A)$ forks over $A$.
The next natural question is whether there is an example in a simple theory. In simple theories, no type in $S(A)$ forks over $A$, but the equivalence between non-forking and Lascar invariance no longer holds. This question was asked by Chernikov and Kaplan in their paper Forking and Dividing in $\mathrm{NTP}_2$ Theories (p. 19 (1)). Chernikov later gave an example in this blog post. I'll describe the example here for completeness.

Recall that a $k$-uniform hypergraph is a set $X$ together with a family of subsets of $X$, called edges, each of which has size $k$. We view a $k$-uniform hypergraph as a structure $(X,R)$, where $R$ is a $k$-ary relation symbol, and $R(a,b,c)$ if and only if $\{a,b,c\}$ is an edge.
A two graph is a $3$-uniform hypergraph (yes, it's terrible terminology) such that for any four distinct points $a,b,c,d$, the number of edges from the set $\{a,b,c,d\}$ is even (out of four possible edges).
The class of finite two graphs is a Fraïssé class, and it turns out that the Fraïssé limit of this class is a reduct of the random graph. Indeed, let $(G,E)$ be the random graph. Define a ternary relation $R$ on $G$ by $R(x,y,z)$ if and only if $x$, $y$, and $z$ are distinct, and the number of $E$-edges from the set $\{x,y,z\}$ is odd (out of three possible edges). Then $(G,R)$ is isomorphic to the Fraïssé limit of finite two graphs. It follows that $T = \mathrm{Th}(G,R)$ is simple.
Now we'll show that relative to $T$, there is no global type which is Lascar invariant over the empty set.
First we show that if $a\neq b$ and $a'\neq b'$, then $ab$ and $a'b'$ have the same Lascar strong type over the empty set. I think the easiest way to see this is by the characterization that "same Lascar strong type over the empty set" is the transitive closure of the relation $c\sim c'$ if $c$ and $c'$ start an indiscernible sequence. Now you can check that if $c,d,c',d'$ are distinct, then $cd\sim c'd'$. Picking any $a''$ and $b''$ distinct from $a,b,a',b'$, we have $ab\sim a''b'' \sim a'b'$, so $ab$ and $a'b'$ have the same Lascar strong type.
Now suppose $p(x)$ is a global type Lascar invariant over the empty set. It follows that either (1) for any $a\neq b$, $R(x,a,b)\in p(x)$, or (2) for any $a\neq b$, $\lnot R(x,a,b)\in p(x)$. In case (1), let $a,b,c$ be such that $\lnot R(a,b,c)$, and let $d$ be a realization of $p$. Then there are three edges from the set $\{a,b,c,d\}$, contradiction. In case (2), let $a,b,c$ be such that $R(a,b,c)$, and let $d$ be a realization of $p$. Then there is one edge from the set $\{a,b,c,d\}$, contradiction.
