For the quadratic equation $6x^2-11x+\alpha=0,$ the roots will be rational if and only if the discriminant is a perfect square.

Compute the discriminant: $D=b^2-4ac=(-11)^2-4(6)(\alpha)=121-24\alpha.$

For rational roots, we need $121-24\alpha$ to be a non-negative perfect square.

Let $121-24\alpha = k^2,$ where $k$ is a non-negative integer.

Since $\alpha$ is a positive integer, $121-24\alpha \ge 0 \implies \alpha \le \frac{121}{24}.$ So possible positive integers are $\alpha=1,2,3,4,5.$

Now check the discriminant for these values:

• For $\alpha=1$: $D=121-24=97,$ not a perfect square.

• For $\alpha=2$: $D=121-48=73,$ not a perfect square.

• For $\alpha=3$: $D=121-72=49=7^2,$ perfect square.

• For $\alpha=4$: $D=121-96=25=5^2,$ perfect square.

• For $\alpha=5$: $D=121-120=1=1^2,$ perfect square.

Thus, the positive integral values of $\alpha$ are: $\alpha=3,4,5.$ Total number of values = $3$.

Hence the correct option is: $\boxed{\text{A: }3}$