Additive but not homogeneous continuous system?
You are misinterpreting the issues entirely. 


  A system with complex inputs and complex outputs that maps its input
  $x(t)$ to output $[x(t)]^*$, the complex
  conjugate of $x(t)$, is additive, is not homogeneous for complex scalars and is homogeneous for real scalars.


Consider the system with complex inputs and complex outputs that maps its
input to its complex conjugate.  Thus, it maps
$$x_1(t) \to [x_1(t)]^*, \quad x_2(t)\to [x_2(t)]^*.$$
So, what happens if the input is the sum $x_1(t)+x_2(t)$? well, the 
rule is that the input gets mapped to its complex conjugate and so
$$x_1(t) + x_2(t) \to [x_1(t) + x_2(t)]^* = [x_1(t)]^* + [x_2(t)]^*.$$
Since this holds for arbitrary choice of $x_1(t)$ and $x_2(t)$, we
conclude that the system is indeed additive: the response to a
sum of signals is the sum of the responses.

Let $\alpha$ denote a complex
number. Then, this "complex conjugator" system maps
$$\alpha\cdot x(t) \to [\alpha\cdot x(t)]^* = [\alpha]^*\cdot [x(t)]^* 
\neq \alpha\cdot [x(t)]^*$$
unless $\alpha$ happens to be a real number.  So, the system is not
homogeneous since the required condition does not hold for arbitrary
(complex) $\alpha$.
However, it is true that if $\alpha$ is restricted to be a real number,
then the scaling property (a.k.a. homogeneity) does hold.



Addendum:  The OP wants


  a system that maps strictly real inputs to strictly real outputs, and that cannot be scaled with a complex (only with real) constants, and that is additive, but not homogeneous. That's my initial question,....


Since the input and output are required to be real, and not complex, the 
notion of "cannot be scaled with a complex constant" does not make sense
any more because the scaling of the input makes it a complex-valued signal, 
and the system allows only for a real-valued input. So essentially the question
boils down to 

"Does there exist a system with real-valued inputs and outputs
that has the additive property but not the homogeneity property?"

to which the answer is No, as pointed out in the 
Wikipedia article on linear systems. Most of the ideas there are reproduced
in the dsp.SE answer
mentioned in a comment by MBaz but the dsp.SE answer goes on to make
the additional claim (not in Wikipedia) that

"not much else is needed to extend to $a \in \mathbb C$"

which is false.  Indeed, as MattL notes carefully in a comment,
MattL's answers are applicable to scaling by real numbers only.

An outline of the proof that for real-valued systems, additivity
implies homogeneity is as follows:

Consider a system with real-valued inputs and outputs that
maps any given $x$ to $A(x)$. The function $A(\cdot)$ has the
property that $A(x_1+x_2) = A(x_1)+A(x_2)$ and the system is called
additive since the response to the sum $x_1+x_2$ is the sum of
the responses $A(x_1)$ and $A(x_2)$.  Mathematical induction extends this to
$$A\left(\sum_{i=1}^n x_i \right) = \sum_{i=1}^n A(x_i).\tag{1}$$


Choosing all the $x_i$ in $(1)$ to have value $0$, we see that
$A(0) = nA(0)$ and so $A(0)=0$.
Choosing all the $x_i$ in $(1)$ to be $x$, we see that $A(nx)=nA(x)$
and so the scaling property holds for positive integers $n$.
Choosing $n=2$ and $x_2 = -x_1$ in $(1)$, we have that
$A(x_1+(-x_1)) = A(x_1)+A(-x_1)$. But $x_1+(-x_1)=0$ and so we
see that $A(0) = 0 = A(x_1)+A(-x_1)$, that is, 
$$A(-x_1)=A((-1)x_1) = -A(x_1) =(-1)A(x_1).$$ So the scaling
property holds for $-1$, and by mathematical induction for
all negative integers as well.
Scaling holds for rational numbers $\displaystyle \frac 1n$ as well. Choose
$\displaystyle x_i = \frac 1n x$ in $(1)$ to get 
$$A(x)=nA\left(\frac 1n x\right) \Rightarrow 
A\left(\frac 1n x\right) = \frac 1n A(x).$$
Mathematical induction allows us to extend this to more general
rational numbers $\displaystyle \frac mn$ as well.
Finally, to show that scaling applies to arbitrary (not necessarily 
rational) real numbers, Wikipedia says that we need to assume that
the mapping is continuous and use the fact the rational numbers
are a dense subset of the reals.


So, except for this last part which requires some mathematics
beyond high school, we can show straightforwardly that
additivity implies homogeneity.