Let $F$ be the center of the division ring $D$. Then $D$ represents its class in the Brauer group $Br(F)$. The opposite ring $D^{opp}$ represents the inverse element. The reason why for example the quaternions are isomorphic to their opposite algebra is that the quaternions are an element of order 2 in $Br(\mathbf{R})$, and hence equal to its own inverse in the Brauer group.
To get a division algebra that is not isomorphic to its opposite algebra we can use an element of order 3 in the Brauer group. One method for constructing those is to start with a (cyclic) Galois extension of number fields $E/F$ such that $[E:F]=3$. Let $\sigma\in Gal(E/F)$ be the generator. Let $\gamma\in F$ be an element that cannot be written in the form $\gamma=N(x)$, where $N:E\rightarrow F, x\mapsto x\sigma(x)\sigma^2(x)$ is the relative norm map. Consider the set of matrices
$$
\mathcal{A}(E,F,\sigma,\gamma)=\left\{
\left(\begin{array}{rrr}
x_0&\sigma(x_2)&\sigma^2(x_1)\\
\gamma x_1&\sigma(x_0)&\sigma^2(x_2)\\
\gamma x_2&\gamma\sigma(x_1)&\sigma^2(x_0)
\end{array}\right)\mid x_0,x_1,x_2\in E\right\}.
$$
A theorem of A. Albert tells us that this forms a division algebra with center $F$, and its order in $Br(F)$ is 3, so it will not be isomorphic to its opposite algebra. The theory is described for example in ch. 8 of Jacobson's Basic Algebra II. The buzzword 'cyclic division algebra' should give you some hits.
For a concrete example consider the following. Let $F=\mathbf{Q}(\sqrt{-3})$ and let
$E=F(\zeta_9)$, with $\zeta_9=e^{2\pi i/9}$. Then $E/F$ is a cubic extension of cyclotomic fields, $\sigma:\zeta_9\mapsto\zeta_9^4$. I claim that the element $2$ does not belong the image of the norm map. This follows from the fact 2 is totally inert in the extension tower $E/F/\mathbf{Q}$. Basically because $GF(2^6)$ is the smallest finite field of characteristic 2 that contains a primitive ninth root of unity. Now, if $2=N(x)$ for some $x\in E$, then 2 must appear as a factor (with a positive coefficient) in the fractional ideal generated by $x$. But the norm map then multiplies that coefficient by 3, and as there were no other primes above 2, we cannot cancel that. Sorry, if this is too sketchy.
Anyway (see Jacobson again), the product of the $\gamma$ elements modulo $N(E^*)$ is the operation in the Brauer group $Br(E/F)\le Br(F).$ Therefore the opposite algebra should correspond to the choice $\gamma=1/2$, (or to the choice $\gamma=4$, as $2\cdot4=N(2)$). So
$$
\mathcal{A}(E,F,\sigma,2)^{opp}\cong
\mathcal{A}(E,F,\sigma,1/2).
$$
and the choices $\gamma=2$ and $\gamma=1/2$ yield non-isomorphic division algebras, as their ratio $=4$ is not in the image of the norm map.
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Edit (added more details here, because another answer links to this answer): In general the cyclic division algebra construction works much the same for any cyclic extension $E/F$. When $[E:F]=n$ we get a set of $n\times n$ matrices with entries in $E$. The number $\gamma$ appears in the lower diagonal part of the matrix. The condition for this to be a division algebra is that $\gamma^k$ should not be a norm for any integer $k, 0<k<n$. Obviously it suffices to check this for (maximal) proper divisors of $n$. In particular, if $n$ is a prime, then it suffices to check that $\gamma$ itself is not a norm.
Edit^2: Matt E's answer here linear algebra over a division ring vs. over a field gives a simpler cyclic division algebra of $3\times3$ matrices with entries in the real subfield of the seventh cyclotomic field.
All you need to do is to produce an Ore domain which isn't commutative and has finite characteristic.
In fact there's an easy way to produce left principal ideal domains (which are necessarily left Ore). The construction is that of the skew-polynomial ring $k[X;\sigma]$, where $\sigma$ is an automorphism of $k$ that isn't the identity. Addition is plain addition of polynomials, but multiplication is governed by the rule $Xa=\sigma(a)X$. When the automorphism isn't the identity, the ring is not commutative.
To achieve this we'll select $k$ to be $F_4$, the field of four elements, and use the Frobenius automorphism $a\mapsto a^2$ (which isn't the identity on $F_4$.)
So now you have $F_4\langle X;\sigma\rangle$, a not-commutative left Ore domain. By Ore's theorem, the classical left ring of fractions exists, and is a division ring that is not commutative, and it's easy to see that it has characteristic $2$.
With regards to the way you were going: there are quaternion-like constructions that do produce division rings from finite fields. You have elements $i,j,k$ such that $i^2=a$, $j^2=b$ and $ij=-ji=k$.
It is possible for this to be a division ring. When the characteristic of $F$ is not $2$, and $-1$ is not the sum of two squares in $F$, then the quaternion construction I described is a division ring (See Corollary 4.23 of K. Conrad's notes.
But also note Corollary 4.24: every quaternion algebra over $F_p$ is isomorphic to $M_2(F_p)$ (hence not a division ring.) So the only hope would be to use more complex fields of finite characteristic.
Best Answer
Wow, three answers and a comment with the same example, and not even the easiest example, IMO! (It's still a good example, though!)
Absolutely! There are lots of nonzero subrings of $\Bbb H$ that would already work. Consider for example $\{a+bi+cj+dk\mid a,b,c,d\in \Bbb Z\}\subseteq\Bbb H$.
It's easy to check that this is a subring of $\Bbb H$, which automatically makes it a domain. Clearly it isn't commutative since $ij\neq ji$. Finally, it doesn't contain $2^{-1}$, so it can't be a division ring.
Absolutely! Actually this will be cheating since it will be the same as my first solution, but you might be interested in what happens anyway.
It turns out that the quaternions can be represented by complex matrices in this way:
$a+bi+cj+dk\mapsto\begin{bmatrix}a+bi&c+di\\-c+di&a-bi\end{bmatrix}\in M_2(\Bbb C)$. So by just taking the image of the ring I described in the first part, you have a subring of $M_2(\Bbb C)$ which is a noncommutative domain, not a division ring.
You could take it even further to real matrices:
$a+bi+cj+dk\mapsto\begin{bmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\\\end{bmatrix}\in M_4(\Bbb R)$ to get another (the same) example. In fact, you can look upon the final example as using a matrix representation of elements of $\Bbb C$: $a+bi\mapsto\begin{bmatrix}a&b\\-b&a\end{bmatrix}$ and plugging that representation into the second example. All three are interlinked.
These are both matrix representations of the original example I suggested.
For a final exotic example that you might be interested in, check out the Hurwitz quaternions. They are an interesting subdomain of $\Bbb H$.