Under some additional assumptions one can prove that $dd^c \max (u,v) \geq \min (dd^cu,dd^cv)$ in $\Omega$. The proof seems long-winded for such a simple statement, but this is the best I can think of now. First of all, I will assume that $u$ and $v$ are continuous in $\Omega \subset \mathbb{C}^n$ and let $w:=\max(u,v)$.
I will also use a criterion of positivity of currents of degree $(n-p,n-p), \quad 0 \leq p \leq n$ in $\Omega$. Let $L$ be a complex linear subspace of $\mathbb{C}^n$ of dimension $p$ and let $g \in \rm U(n)$ be such that $g(L)$ is defined by equations $u_{p+1}=...=u_n=0$. Define a $(p,p)$-form $\tau(L):= g^*\hat{\beta}$, where $\hat{\beta}=(i/2)du_1\wedge d\bar{u}_1\wedge...\wedge (i/2)du_p\wedge d\bar{u}_p$. This does not depend on the choice of orthonormal coordinates and gives a positive form. Now, Proposition 2.10 in: P. Lelong, L. Gruman: Entire functions of several complex variables, Springer-Verlag 1986 says the following: A $(n-p,n-p)$-current $T$ in $\Omega$ is positive if and only if for every linear subspace $L$ of dimension $p$ the current $T \wedge \tau(L)$ is a positive distribution (and hence a measure).
Because of linearity of $dd^c$ and of the equality $dd^ch \wedge S = dd^c(hS)$, which holds when u is plurisubharmonic and locally integrable with respect to the mass of $S$, it is enough to prove that $dd^c w \wedge \tau(L) \geq \min(dd^cu,dd^cv) \wedge \tau(L)$ for every linear subspace $L$ of dimension $n-1$.
Here is the proof, modeled on Proposition 2.8 in: Bedford, Eric; Taylor, B. A.: The Dirichlet problem for a complex Monge-Amp`{e}re equation. \emph{Invent. Math. 37} (1976), no. 1, 1-44. The authors of this paper study Monge-Ampere measures, not trace measures, but the argument goes through. I cannot resist reproducing it here in full, partly because of their cute application of the fact that an uncountable sum of positive real numbers is infinite (for the proof of it, see e.g. http://www.xamuel.com/uncountable-sums/; there is also related discussion in
http://math.stackexchange.com/questions/70194/does-uncountable-summation-with-a-finite-sum-ever-occur-in-mathematics).

Claim: Let $\Omega,u,v,w$ be as above. If for every $L$ of complex dimension $n-1$ the measure $\mu_L := \min(dd^cu, dd^cv) \wedge \tau(L)$ is finite, then $dd^c w \wedge \tau(L)\geq \min(dd^cu, dd^cv)\wedge \tau(L)=\mu_L$.
Proof: Consider first the case when $\mu_L(\{u=v\})=0$.
Let $\Omega_1 = \{u<v\}$,$\Omega_2=\{v<u\}$. Then $dd^cw \wedge \tau(L)=dd^cv \wedge \tau(L)\geq \mu_L$ in $\Omega_1$ and similarly $dd^cw \geq \min(dd^cu, dd^cv)$ in $\Omega_2$. And of course $dd^cw \wedge \tau(L)(\{u=v\}) \geq 0 = \mu_L(\{u=v\})$.\
Now let us consider $L$ such that $\mu_L(\{u=v\})> 0$. Then replacing $v$ by $v+\varepsilon$ for small $\varepsilon$ we still have $\mu_L = \min(dd^cu, dd^c(v+\varepsilon)) \wedge \tau(L)$ and arguing as above we can prove that $dd^c \max(u,v+\varepsilon) \wedge \tau(L) \geq \mu_L$ for all $\varepsilon$ for which $\mu_L(\{z\in \mathbb{C}^n: u(z)=v(z)+\varepsilon \})=0$. This happens for all but countably many $\varepsilon$ , which allows us to obtain the inequality in the general case by letting $\varepsilon \to 0$ in a way that avoids those countably many values.

I do not know if and how the assumption of continuity can be relaxed. Bedford and Taylor give a counterexample of discontinuous plurisubharmonic functions for which the inequality $(dd^cw)^n \geq \min((dd^cu)^n,(dd^cv)^n)$ does not hold.
If one wants to know which currents satisfy the assumption of finiteness of the trace measures, one should look at
Fornaess, John Erik; Sibony, Nessim: Oka's inequality for currents and applications. \emph{Math. Ann. 301} (1995), no. 3, 399-419, where similar estimates were investigated.