A real-valued function has to output a real number or a vector that is composed only of real numbers.
But can a scalar-valued function or vector-valued function output an imaginary number if it is not also labeled as real-valued?
For example, is this a scalar valued function?
$$ f:x \mapsto x\sqrt{-1} $$
On
Scalar means with values in the base field, whatever that field is. It depends on the context. The terminology is derived from the fact that when you multiply a set of vectors by a number, you scale the picture (you change the scale but not the overall shape). Hence numbers are "scalers".
When geometry and linear algebra get axiomatised, and you start working over an arbitrary vector/affine space over some base field, scalars are elements of the field. In particular, it could very well be the field of complex numbers.
A general truth in mathematics is that functions do not exist independently from their domain and codomain. This is a misconception generated by those pre-calc problems where they asked you to "find the domain" of a certain formula. So $f$ by itself (the "mapping rule") is not a function; $f : A \to B$, instead, is. Therefore, you need to specify what $A$ and $B$ are when defining a function.
Real numbers live in $\mathbb R$. Real euclidean vectors live in $\mathbb R^n$ for some $n$. A function from $f : \mathbb R \to \mathbb R$ is called real(-valued) function of one real variable. A function $f : \mathbb R \to \mathbb R^n$ is called vector(-valued) function of one real variable. A function $f : \mathbb R^k \to \mathbb R$ is called a real(-valued) function of $k$ real variables. Just guess what functions like $f : \mathbb R^k \to \mathbb R^n$ are.
In the context of real vector spaces, and in particular $\mathbb R^n$, scalars are the elements of the base field of the vector space. In the case of $\mathbb R^n$, that field is most naturally $\mathbb R$. So any time you find "real(-valued)" in the above nomenclatures, you may substitute scalar(-valued).
As you know, the imaginary unit $i$ is not in $\mathbb R$, so the function you proposed (wherever your $x$ comes from) may never be of the kinds listed above. However, if $x$ came from $\mathbb C$ or $\mathbb C^n$, the map $x \mapsto ix$ would not be such a strange sight.