If you had a "free division ring" $F$ on a set $X$, any division ring $R$, then any set theoretic map $X\to R$ would correspond to a unique division ring homomorphism $F\to R$. If $X$ has at least two elements $x\neq y$, let $R=\mathbb{Q}$. Then you cannot extend both a set theoretic map $f\colon X\to R$ that sends $x$ to $0$ and $y$ to $1$, and a set-theoretic map $g\colon X\to R$ that sends $x$ to $1$ and $y$ to $0$: division rings are simple, so any homomorphism must be either one-to-one or the zero map. (I put both maps, in case one wonders whether you can have $x$ correspond to the zero element of $F$). So, no, you cannot have "free division rings", much like you cannot have "free fields".

If you had a "free division ring" $F$ on a set $X$, and any division ring $R$, then any set theoretic map $X\to R$ would correspond to a unique division ring homomorphism $F\to R$. If $X$ has at least two elements $x\neq y$, let $R=\mathbb{Q}$. Then you cannot extend both a set theoretic map $f\colon X\to R$ that sends $x$ to $0$ and $y$ to $1$, and a set-theoretic map $g\colon X\to R$ that sends $x$ to $1$ and $y$ to $0$: division rings are simple, so any homomorphism must be either one-to-one or the zero map. (I put both maps, in case one wonders whether you can have $x$ correspond to the zero element of $F$). So, no, you cannot have "free division rings", much like you cannot have "free fields".