A measure algebra is a pair $(\frak{A},\mu)$ where $\frak{A}$ is a Boolean $\sigma$-algebra and $\mu$ is a strictly positive measure $\mu$ on $\frak{A}$.
Can $\frak{A}$ in the above definition be a free Boolean $\sigma$-algebra?
I would guess yes because a free Boolean $\sigma$-algebra is a Boolean $\sigma$-algebra. Yet, I remain very confused about free Boolean algebras. In other words, can I always define a measure on a free Boolean $\sigma$-algebra without encountering any problem?
In the definition of a measure algebra, all that is required of $\mathfrak{A}$ is that it is a Boolean $\sigma$-algebra. So it can be any Boolean $\sigma$-algebra, including a free Boolean $\sigma$-algebra, as long as $\mu$ is a strictly positive measure on it.
However, if $\mathfrak{A}$ is a free Boolean $\sigma$-algebra on infinitely many generators, then there does not actually exist any strictly positive measure on it, so there is no choice of $\mu$ that will satisfy the definition. To prove this, suppose $\mathfrak{A}$ is free on an infinite set $X$ and choose distinct elements $x_n\in X$ for each $n\in\mathbb{N}$. For each subset $A\subseteq\mathbb{N}$, define an element $x_A\in\mathfrak{A}$ by $$x_A=\bigwedge_{n\in A}x_n\wedge\bigwedge_{n\not\in A}\neg x_n.$$ Note that if $A\neq B$, then $x_A\wedge x_B=0$, since $A$ and $B$ disagree on some $n\in\mathbb{N}$ and so $x_n$ appears in one of $x_A$ and $x_B$ and $\neg x_n$ appears in the other. Also, note that $x_A\neq 0$ for all $A$: since $\mathfrak{A}$ is free on $X$ there is a $\sigma$-homomorphism $f:\mathfrak{A}\to\{0,1\}$ sending $x_n$ to $1$ for all $n\in A$ and sending all other elements of $X$ to $0$, and then $f(x_A)=1$.
Now suppose $\mu$ is a strictly positive measure on $\mathfrak{A}$. Then $\mu(x_A)>0$ for all $A\subseteq\mathbb{N}$. Since there are uncountably many such $A$, there is some $n\in\mathbb{N}$ such that $\mu(x_A)>1/n$ for infinitely many different $A$. But these $x_A$ are disjoint, so if you take the join of more than $n$ of them you get an element of $\mathfrak{A}$ whose measure is greater than $1$, which is a contradiction. Thus no such $\mu$ can exist.