Let $F(p,q,r,s)=p^6-q^6+r^3-s^3$. I prove that there are polynomials $p, q, r, s\in\mathbb{Q}[t]$ such that $deg_{t}(F(p(t),q(t),r(t),s(t)))=1$. This is enough to get the result. Indeed, if $F(p(t),q(t),r(t),s(t))=ut+v$, then taking $t=n/u-v$ we will get representation of $n$ by $F$ (with rational entries). 

To find required polynomials, we assume that they are in the following form
$$
p=t+a,\quad q=t, \quad r=bt^2+ct+d, \quad s=bt^2+et+f, 
$$
for suitable choosen $a, b, c, d, e, f$. Subsituting this into our polynomial $F$, we get
$$
F(p, q, r, s)=\sum_{i=0}^{5}a_{i}t^{i}.
$$ We thus find that the necessary condition for $deg(F(p, q, r, s))\leq 1$ is the existence of the solution of the system 
$$
S:\;a_{2}=a_{3}=a_{4}=a_{5}=0.
$$ 
More precisely, we have
\begin{align*}
a_{2}&=3 \left(5 a^4+b d^2-b f^2+c^2 d-e^2 f\right),\\
a_{3}&=20 a^3+6 b c d-6 b e f+c^3-e^3,\\
a_{4}&=3 \left(5a^2+b^2 d-b^2 f+b c^2-b e^2\right),\\
a_{5}&=3 \left(2 a+b^2 c-b^2 e\right).
\end{align*} 
The subsystem $S':\;a_{3}=a_{4}=a_{5}=0$ is triangular with respect to the variables $d, e, f$ and we found that
\begin{align*}
d&=\frac{10 a^2 b^6-30 a^2 b^3+20 a^2-15 a b^5 c+30 a b^2 c+9 b^4 c^2}{6 b^5},\\
e&=\frac{2 a+b^2c}{b^2},\\
f&=\frac{10 a^2 b^6-4 a^2-15 a b^5 c+6 a b^2 c+9 b^4 c^2}{6 b^5}.
\end{align*}
Putting the computed values of $d, e, f$ into the last equation $a_{2}=0$ we get, after clearing denominator and constants, the equation
$$
a \left(a b^3-a-b^2 c\right) \left(7 a^2 b^6-12 a^2 b^3+8 a^2-12 a b^5 c+12 a b^2 c+6 b^4c^2\right)=0.
$$
We are lucky here because the second factor lead to the solution we are looking for. More precisely, we take 
$$
c=\frac{a (b-1) \left(b^2+b+1\right)}{b^2}.
$$
Tracing back our resoning, we obtain solution of our system $S$ and for most choices of $a, b$, the corresponding values of $a_{0}, a_{1}$ are nonzero (if $a_{1}=0$ we loose). The full expansions of $p, q, r, s$ are given at the end. To get positive values of $p, q, r, s$ we need to manipulate with $a, b$. To get an example take $a=-1/2, b=2$ and then we get the identity
$$
p^6-q^6+r^3-s^3=n,
$$   
where
\begin{align*}
p&=\frac{1048576 n-5355}{21420},\\
q&=\frac{1048576 n+5355}{21420},\\
r&=\frac{35184372088832n^2+44920995840 n+1519829325}{7341062400},\\
s&=\frac{35184372088832 n^2-44920995840n+1519829325}{7341062400}
\end{align*}
Note that if $n>6/1000$ then $p, q, r, s>0$. Taking $a$ small enough and negative we can cover all positive $n$. Similarly, by taking $a$ positive and small enough (and $b=2$) we can cover all negative values of $n$. 

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Full expansion of $p, q, r, s$ takes the form
\begin{align*}
p&=\frac{a}{2}-\frac{6b^{12}}{D}n,\\
q&=-\frac{a}{2}-\frac{6b^{12}}{D}n,\\
r&=\frac{a^2 \left(5 b^6-2\right)}{12 b^5}-\frac{6 a b^{10}}{D}n+\frac{36 b^{25}}{D^2}, \\
s&=\frac{a^2 \left(5 b^6-2\right)}{12 b^5}+\frac{6 a b^{10}}{D}n+\frac{36 b^{25}}{D^2},
\end{align*}
where
$$
D=a^5 \left(b^3-1\right) \left(b^3+1\right) \left(2 b^3-1\right) \left(2 b^3+1\right).
$$