Here is a slightly ridiculous solution. We'll use three well-known formulas for the area of $\triangle ABC$: $$[ABC]=\frac{abc}{4R}=\frac{ab\sin C}{2}=\sqrt{s(s-a)(s-b)(s-c)}.$$First of all, because $a,b,c,R$ are all integers, we know that $[ABC]$ is rational. But then $$4[ABC]=\sqrt{2s(2s-2a)(2s-2b)(2s-2c)}\in\mathbb{Z}$$(since $2s(2s-2a)(2s-2b)(2s-2c)$ is an integer, its square root, if it is rational, must be an integer) so $[ABC]=n/4$ for some $n\in\mathbb{Z}.$ Then $n=abc/R$ is an integer, so $R$ divides one of $a,b,c$; WLOG $a=kR$ for some $k\in\mathbb{Z}.$ But now $$\frac{abc}{4R}=\frac{bc\sin A}{2}\implies\sin A=\frac{a}{2R}=\frac{k}{2}$$and since $\sin A\in[0,1]$ we have $k\in\{1,2\}.$ If $k=1$ then we have from Law of Cosines $$a^2=R^2=b^2+c^2-2bc\cos A=b^2+c^2-bc\sqrt{3},$$which is absurd because $R,b,c$ are all integers. So $k=2\implies\sin A=1\implies A=90^{\circ}$ and we are done.

BobThePotato wrote: Here is a slightly ridiculous solution. We'll use three well-known formulas for the area of $\triangle ABC$: $$[ABC]=\frac{abc}{4R}=\frac{ab\sin C}{2}=\sqrt{s(s-a)(s-b)(s-c)}.$$First of all, because $a,b,c,R$ are all integers, we know that $[ABC]$ is rational. But then $$4[ABC]=\sqrt{2s(2s-2a)(2s-2b)(2s-2c)}\in\mathbb{Z}$$(since $2s(2s-2a)(2s-2b)(2s-2c)$ is an integer, its square root, if it is rational, must be an integer) so $[ABC]=n/4$ for some $n\in\mathbb{Z}.$ Then $n=abc/R$ is an integer, so $R$ divides one of $a,b,c$; WLOG $a=kR$ for some $k\in\mathbb{Z}.$ But now $$\frac{abc}{4R}=\frac{bc\sin A}{2}\implies\sin A=\frac{a}{2R}=\frac{k}{2}$$and since $\sin A\in[0,1]$ we have $k\in\{1,2\}.$ If $k=1$ then we have from Law of Cosines $$a^2=R^2=b^2+c^2-2bc\cos A=b^2+c^2-bc\sqrt{3},$$which is absurd because $R,b,c$ are all integers. So $k=2\implies\sin A=1\implies A=90^{\circ}$ and we are done. The angles in this triangle are not necessarily rational. For example, the 6,8,10 pythagorean triangle is a solution as it has outradius 5, but it does not have rational angles.

I never claimed that the angles themselves are rational aside from $A.$ Rather, the sines of the angles are necessarily rational by the given conditions, and one of the sines ($\sin A$ in the solution above) must be either $1/2,$ which implies $A=30^\circ,150^\circ$ (though both were discounted), or $1,$ which implies $A=90^\circ.$