In Peskin & Schroder, Introduction to Quantum Field Theory equation (15.82) states that
$$ t^a_{\bar{r}} = -(t^a_{r})^* = (t^a_{r})^T $$
Why is the representation which satisfies
$$ t_{\overline{r}}^{a}=Ut_{r}^{a}U^{\dagger} $$
called real?
Does this real means something physically real?
This means mathematically real, that is, a real-valued representation.
Have you studied the Representation Theory of (Lie) Groups and of Lie Algebras?
A representation may be real, in which the matrices that represent the linear action of the group element are matrices with real-valued elements. We than say that the matrices are in some subgroup of $GL(n,\mathbb{R})$.
Converserly, a representation may be Complex-Valued, and have matrices which are in some subgroup of $GL(n,\mathbb{C})$.
In fact, if you had looked at equation (15.82) in Peskin & Schroeder, which is defining the Conjugate Representation $\bar{r}$ or some representation $r$
$$ t^a_{\bar{r}} = - (t^a_{r})^* = - (t^a_{r})T $$
you would see in the following paragraph he explains his definition of a real representation $r$ to be one for which
$$ r \equiv \bar{r} $$
That is, for $r$ to be real, there must be a unitary transformation
$$ t^a_{\bar{r}} = U t^a_r U^{\dagger} $$
as you say, and the complex-conjugate operation above is trivial. Equivalently, $r$ must be real-valued.
