An $R$-algebra $A$ being split means that ${\rm End}_A M = R$ for any simple $A$-module $M$, to the best of my knowledge. In any case, statement is false though. If any of the $A_i$ are not distinct, then we can twist the bilinear form a little, while preserving the desired properties, and end up with a counterexample. An explanation follows, but it would be worth considering it yourself first with say, the case $A=S\oplus S$ for some simple $R$-algebra $S$.
Take for example $A=Re_1\oplus Re_2$ to be the semisimple $R$-algebra, where $e_1,e_2$ are orthogonal idempotents.
This is split because the simple $A$-modules are just $e_1A$ and $e_2A$ and so it is apparent that the endomorphisms of these are just multiples of the identity.
Now define a bilinear form on $A$ by $(x,y)=x_1y_2+x_2y_1$ where $x=x_1e_1+x_2e_2$ and $y=y_1e_1+y_2e_2$ with $x_1,x_2,y_1,y_2\in R$. Then this bilinear form is symmetric and nondegenerate since $(x,e_2)=x_1$ and $(x,e_1)=x_2$. By construction, the simple subalgebras are isotropic (i.e. $(e_i,e_i)=0$) but the inner product between them is nonzero (i.e. $(e_1,e_2)=1$).
By the way, a more concrete way to see all of this is that this inner product is given by $\begin{bmatrix}x_1 & x_2\end{bmatrix}\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}y_1 \\ y_2\end{bmatrix},$ and nondegeneracy and symmetry follow from the fact the matrixis invertible and symmetric.