The answer is in the negative. 

Let $f$ and $g$ be two upper densities (in the sense of the OP), and let $\alpha \in [0,1]$ and $q \in [1,\infty[$. Then the function 
$$h: \mathcal P(\mathbf N^+) \to \mathbf R: X \mapsto (\alpha (f(X))^q + (1-\alpha) (g(X))^q)^{\frac{1}{q}}$$ is an upper density too (in particular, condition (F3) follows from Minkowski's inequality, which is why we need $q \ge 1$). 

Next, fix a set $X \subseteq 2\cdot\mathbf N^+$, and suppose to a contradiction that $h$ is ``weakly additive'' (that is, $f(A \cup B) = f(A) + f(B)$ for all disjoint $A, B \subseteq \mathbf N^+$ such that $B$ is an (infinite) arithmetic progression), regardless of the actual values of the parameters $\alpha$ and $q$. Then, also $f$ and $g$ are weakly additive, and using that $f(2\cdot\mathbf N^+ + 1) = g(2\cdot\mathbf N^+ + 1)=\frac{1}{2}$, we obtain 
$$
(\alpha (x+1)^q + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (\alpha x^q + (1-\alpha)y^q)^{\frac{1}{q}} + 1,
$$ 
where $x := 2f(X)$ and $y := 2g(X)$.

On the other hand, an appropriate choice of $f$, $g$ and $X$ makes it possible to have $x$ equal to zero while $y$ takes any value in the interval $[0,1]$: This can be achieved, for instance, by letting $f$ be the upper asymptotic density (on $\mathbf N^+$), $g$ the upper Banach density, and $X$ a suitable subset of the intersection, $S$, of $\bigcup_{n \ge 1} [\![2^n, 2^n + n]\!]$ and $2 \cdot\mathbf N^+$, and by considering that (i) the upper Banach density of $S$ is $\frac{1}{2}$, (ii) the upper asymptotic and upper Banach densities are upper densities, and (iii) upper densities have the strong, and hence the weak, Darboux property (by the main theorem [here][1]).

Accordingly, we should have
$$(\alpha + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (1-\alpha)^{\frac{1}{q}}y + 1$$
for all $\alpha, y \in [0,1]$ and $q \in [1,\infty[$, which, however, is blatantly false. []

**Added later.** If you assume $\alpha = \frac{1}{2}$  and $q = 2$ in the last displayed equation, you don't even need to know that the upper Banach density has the weak Darboux property, since then you end up with the equation 
$$\sqrt{1 + (1+y)^2} = y + \sqrt{2},$$
which has a unique solution for $y \in \bf R$ (namely, $y = 0$).

  [1]: http://arxiv.org/abs/1510.07473