Calculating the molar iodine concentration of a CT contrast agent
I highly recommend not looking for literature from non-authoritative sources such as drugs.com: that's a paid-advertising site and is full of partial/misinformation.

For your compound, see the paperwork from the people who make it - Bracco Imaging. 

Each $\mathrm{mL}$ of Isovue-300 contains more than what you list (namely, water, sodium hydroxide, and hydrochloric acid). That means your approach is close. 

The wording about the 300 $\mathrm{mg}$ of organically-bound $\ce{I}$ per $\mathrm{mL}$ of solution makes sense when you consider the atomic weight of $\ce{I}$ (about 127 $\mathrm{mg}\,\cdot\mathrm{mmol}^{-1}$) and that 3 atoms of iodine comes out to 381 $\mathrm{mg}$. Thus, for that slightly dilute solution, the 300 $\mathrm{mg}$ makes sense. It's not an additional 300 - it's what's in the $\mathrm{mL}$ as-is.

So, the molarity of $\ce{I}$ is (I abbreviate iopamidol as "iop" below to save space):

$$\left({612\,\mathrm{mg\,iop}\over 1\,\mathrm{mL}}\right)\cdot
\left({3\,\mathrm{mmol}\,\ce{I}\over 777.09\,\mathrm{mg\,iop}}\right)\cdot
\left({1\,\mathrm{mol}\,\ce{I}\over 1000\,\mathrm{mmol}\,\ce{I}}\right) \cdot
\left({1000\,\mathrm{mL}\over 1\,\mathrm{L}}\right)
= 2.36\,\mathrm{M}$$