Here's what you can do. Let $A=(a_{ij})_{1\le i\le n, 1\le j\le m}$ be your matrix. I assume that all the entries are in some (not necessarily f.d.) extension field of $\Bbb{Q}$.
Let $V_j$ be the $\Bbb{Q}$-span of all the entries $a_{ij}, 1\le i\le n$. That is, the span of the entries on column $j$. Note that the space $V_j$ is finitely generated. Therefore we can find a basis $\mathcal{B}_j$ of $V_j$ over $\Bbb{Q}$. We can now obviously write all the entries $a_{ij}$, $1\le i\le n$, as $\Bbb{Q}$-linear combinations of numbers in $\mathcal{B}_j$.
Here's how you can answer your question using Gaussian elimination (over $\Bbb{Q}$):
- Replace column $j$ of $A$ with $|\mathcal{B}_j|$ columns simply by replacing each entry $a_{ij}$ with its vector of coordinates with respect to $\mathcal{B}_j$. Call the resulting matrix $\tilde{A}$.
- Perform the usual Gaussian elimination on $\tilde{A}$. Observe that as all the entries of $\tilde{A}$ are in $\Bbb{Q}$, this process will never take you outside of $\Bbb{Q}$.
- You can then read the row rank over $\Bbb{Q}$ of $A$ as the rank of $\tilde{A}$. This is because the bases $\mathcal{B}_j$ faithfully represent linear (in)dependencies over $\Bbb{Q}$. Similarly, you can find a basis for the row space of $A$ over $\Bbb{Q}$ simply by rewriting the $j$th blocks of the non-zero rows of the (reduced) row echelon form of $\tilde{A}$ as elements of $V_j$.
- As usual, the rank of $\tilde{A}$ is also equal to the size of its largest non-vanishing minor.
As a first example consider the one I proffered in the comments (to calibrate my understanding of the question):
$$
A=\left(\begin{array}{c}1\\ \sqrt{2}\\ \sqrt{3}\\ \sqrt{6} \end{array}\right).
$$
There is a single column. The space $V_1$ is the degree four extension field $F=\Bbb{Q}(\sqrt2,\sqrt3)$. From the first course on field extensions we know that
$\mathcal{B}_1=\{1,\sqrt2,\sqrt3,\sqrt6\}$ is a $\Bbb{Q}$-basis of $F$. Writing
the elements of $A$ in terms of this basis thus leads to the matrix
$$
\tilde{A}=\left(\begin{array}{rrrr}1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right).
$$
We immediately see that $\tilde{A}$ is already in the row echelon form. Therefore $\tilde{A}$ has full rank, and we can conclude that the row rank of $A$ over $\Bbb{Q}$ is four.
As another example let's consider the matrix
$$
B=\left(\begin{array}{rr}
1&1\\
\sqrt2&i\\
3&1+i\\
2+\sqrt2&2i\end{array}\right).
$$
Here the entries on the first column are in the span of $\mathcal{B}_1=\{1,\sqrt2\}$
while the entries on the second column are in the span of $\mathcal{B}_2=\{1,i\}$.
Replacing the entries of $B$ with their respective coordinate vectors gives us the matrix
$$
\tilde{B}=\left(\begin{array}{rrrr}
1&0&1&0\\
0&1&0&1\\
3&0&1&1\\
2&1&0&2
\end{array}\right).
$$
Subtracting the appropriate multiples of the first row from the others gives
$$
\rightarrow
\left(\begin{array}{rrrr}
1&0&1&0\\
0&1&0&1\\
0&0&-2&1\\
0&1&-2&2
\end{array}\right).
$$
Further subracting the prescribed multiples of the second row from those below it yields
$$
\rightarrow
\left(\begin{array}{rrrr}
1&0&1&0\\
0&1&0&1\\
0&0&-2&1\\
0&0&-2&1
\end{array}\right).
$$
From here we see that the last row is a replica of the third, and will vanish in the next step. We can conclude that $\tilde{B}$ has rank $3$, and therefore also the the row rank of $B$ over $\Bbb{Q}$ is also $3$. Indeed, if we denote by $u_i$ the $i$th row of $B$, we easily see that $u_3-u_1=(2,i)=u_4-u_2,$ so the rows of $B$ are linearly dependent over $\Bbb{Q}$.
Running this algorithm will make it necessary to correctly identify the dimensions of the spaces $V_j$ (after which finding bases for them is usually straightforward). Some basic facts about algebraic extensions suffice to handle my example cases, but I do not know, whether such techniques will be available in the examples that interest you the most.
In the second example from the question the space $V_1$ consists of degree one polynomials with coefficients from $\Bbb{Q}$, so $\mathcal{B}_1=\{1,x\}$.
We see that $V_2=\sqrt{2}V_1$, so we can use $\mathcal{B}_2=\sqrt2\mathcal{B}_1$. Finally we see that $V_3$ is the 1-dimensional space spanned by $e^x$. For this matrix, call it $C$, we thus get
$$
\tilde{C}=\left(\begin{array}{rrrrr}
1&1&0&1&1\\
1&0&1&0&1\\
0&0&0&0&1\end{array}\right).
$$
Looking at columns $2,3$ and $5$ we immediately see that $\tilde{C}$ has full rank. Therefore the row rank of $C$ over $\Bbb{Q}$ is also $3$.
Best Answer
In general if you extend field $F$ as $F[\sqrt p]$, then you can construct inverse by doing the following: $$ \frac1{a+b\sqrt p} = \frac{a-b\sqrt{p}}{a^2-b^2p} = (a^2-b^2p)^{-1}(a-b\sqrt p). $$
You can consider $Q[\sqrt 2, \sqrt 3]=Q[\sqrt 2][\sqrt 3]$: $$ \frac1{E+F \sqrt2+G \sqrt3 + H \sqrt6} = \frac1{(E+F\sqrt2)+(G+H\sqrt2)\sqrt3} = \frac{(E+F\sqrt2)-(G+H\sqrt2)\sqrt3}{(E+F\sqrt2)^2-3(G+H\sqrt2)^2} $$
Now you have to invert the denominator from only $Q[\sqrt 2]$.