The answer is no, on the assumption that any process of "simplification" will yield a polynomial in $x$ with rational coefficients if the coefficients of the given quadratics are rational.
Consider any case where $h(x) = m(x)$ and $f(x) = 0$.
If $x$ satisfies the given equation, then
$$
\sqrt{g(x)} - \sqrt{l(x)} = k(x)
$$
and therefore
$$
\tag{$*$}\label{eq:1}
4g(x)l(x) = (k(x)^2 - g(x) - l(x))^2.
$$
Now specialise further, to the case:
\begin{align*}
g(x) & = x^2 + \tfrac12, \\
l(x) & = x^2 - \tfrac12, \\
k(x) & = x^2 + 1.
\end{align*}
\eqref{eq:1} becomes
$$
4x^4 - 1 = (x^4 + 1)^2.
$$
This can be solved explicitly in terms of nested square roots, but it is enough for our purposes to rewrite it as
$$
x^8 - 2x^4 + 2 = 0,
$$
and observe that by Eisenstein's criterion the polynomial on the left hand side is irreducible over the rationals, therefore $x$ does not satisfy any polynomial equation of degree less than $8$ with rational coefficients.
Update
This uses a second example to answer the question of whether reduction to a quartic equation is possible. It takes into account the clarification of the question in comments made since this answer was first posted. A further update below uses a third example to answer the more general question asked in the title of the thread: does the equation have an algebraic solution
(which might not be obtained by reduction to a quartic)?
Consider the special case
$$
\frac{x^2 + 1 + \sqrt{x^2 + x + \tfrac12}}{x^2 + 1} =
\frac{2x^2 + \sqrt{x^2 - x + \tfrac12}}{x^2 + 1}.
$$
Here we have
\begin{align*}
g(x) & = x^2 + x + \tfrac12 =
\left(x + \tfrac12\right)^2 + \tfrac14 > 0, \\
l(x) & = x^2 - x + \tfrac12 =
\left(x - \tfrac12\right)^2 + \tfrac14 > 0,
\end{align*}
for all real $x$, so the stated conditions are satisfied. The
equation simplifies to
\begin{equation}
\tag{$**$}\label{3380555:eq:2}
\sqrt{g(x)} - \sqrt{l(x)} = n(x),
\end{equation}
where $n(x) = x^2 - 1$. The four equations
$$
\pm\sqrt{g(x)} \pm\sqrt{l(x)} = n(x)
$$
together are equivalent to a single polynomial equation
$$
4g(x)l(x) = (n(x)^2 - g(x) - l(x))^2.
$$
Because
$g(x)l(x) = \left(x^2 + \tfrac12\right)^2 - x^2 = x^4 + \tfrac14$,
and $g(x) + l(x) = 2x^2 + 1$, the polynomial equation reduces to
$$
4x^4 + 1 = (x^4 - 4x^2)^2,
$$
i.e. $p(x) = 0$, where
$$
p(x) = x^8 - 8x^6 + 12x^4 - 1.
$$
Write this as $p(x) = q(x^2)$, where
$q(y) = y^4 - 8y^3 + 12y^2 - 1 = y^2(y - 2)(y - 6) - 1$.
$$
\begin{array}{r|rrrrrr}
y & -1 & 0 & 1 & 2 & 6 & 7 \\
q(y) & 20 & -1 & 4 & -1 & -1 & 244
\end{array}
$$
$p(x)$ therefore has six real roots and two imaginary roots.
Corresponding to the root of $q(y)$ just below $2$, $p(x)$ has the
real root
$$
x \bumpeq 1.390776704210124
$$
(as calculated in Python using Newton's method).
Numerical calculation confirms that this value of $x$ is also a
solution of \eqref{3380555:eq:2}. (A comment below shows that this is the only positive solution.)
I didn't have the heart to try to solve the quartic equation
$q(y) = 0$ to obtain a closed-form expression for $x$ and try to use
it to prove that \eqref{3380555:eq:2} is satisfied!
Instead one can prove it by using inequalities. We have
$1 < x^2 < 2$, therefore $0 < n(x) < 1$. Also $g(x) > l(x)$, so we
cannot have $-\sqrt{g(x)} \pm\sqrt{l(x)} = n(x)$.
Next, $\sqrt{g(x)} > x + \tfrac12$ and $\sqrt{l(x)} > x - \tfrac12$,
so $\sqrt{g(x)} + \sqrt{l(x)} > 2x > 2 > n(x)$.
Now only \eqref{3380555:eq:2} remains as a possibility.
We have $p(X) \equiv X^8 + X^6 + 2 \pmod3$, and the latter
polynomial is listed as being irreducible on this web site:
Irreducible trinomials over GF(3).
I take this to be good evidence that $p(x)$ is irreducible, although
I haven't verified the result stated on the web site. It follows as
before that $x$ does not satisfy any polynomial equation of degree
less than $8$ with rational coefficients.
Update 2
Consider the special case
$$
\frac{x^2 + 2 + \sqrt{x^2 + x + 1}}{x^2 + 2} =
\frac{x^2 + 1 + \sqrt{x^2 - x + 1}}{x^2 + 1}.
$$
This is equivalent to
$$
(x^2 + 1)\sqrt{x^2 + x + 1} = (x^2 + 2)\sqrt{x^2 - x + 1}.
$$
Because both sides of the latter equation are positive, it is in turn equivalent, for real $x$, to the polynomial equation
$$
(x^2 + 1)^2(x^2 + x + 1) = (x^2 + 2)^2(x^2 - x + 1).
$$
Simplifying:
$$
x[(x^2 + 1)^2 + (x^2 + 2)^2] = (x^2 + 1)[(x^2 + 2)^2 - (x^2 + 1)^2].
$$
This equation simplifies in turn to the quintic equation $p(x) = 0$, where
$$
p(x) = 2x^5 - 2x^4 + 6x^3 - 5x^2 + 5x - 3.
$$
$p'''(x) = 120x^2 - 48x + 36 = 120\left(x-\tfrac15\right)^2 + 31\tfrac15 > 0$
for all $x$, so the cubic $p''(x)$ is strictly increasing. Its unique zero is
$x_0 \bumpeq 0.30861$. But $p'(x_0) \bumpeq 3.4838$, therefore $p'(x) > 0$ for
all $x$, therefore $p(x)$ is strictly increasing, therefore it has a unique real zero:
$$
x \bumpeq 0.705526664475018.
$$
We have $3p(X) \equiv X^5 + 4X^4 + 3X^3 + 1 \pmod{5}$, which is listed as irreducible here, implying that $p(X)$ is irreducible in
$\mathbb{Q}[X]$.
According to GAP 4.8.10:
gap> GaloisType(UnivariatePolynomial(Rationals,[-3,5,-5,6,-2,2]));
5
gap> TransitiveGroup(5,5);
S5
I think this proves that the Galois group of $p(X)$ is isomorphic to $S_5$. This group is famously not solvable. Therefore, no root of $p(X)$, including its unique real root shown above, is given by an expression involving only integer constants, field operations, and/or extractions of roots.
It follows that the general equation
$$
\frac{f(x) + \sqrt{ g(x)}}{h(x)} = \frac{k(x) + \sqrt{ l(x)}}{m(x)}
$$
has no general solution expressed in terms of field operations, extractions of roots, and the coefficients of $f, g, h, k, l, m$.
