If I didn't screw something, I guess: yes, she can! First of all, let's see that in every moment of the game, one can describe full information of the given situation in a pair of numbers $(m,r)$, where $m$ is the last move ie. digit modulo $3$ (so $m\in\{0,1,2\}$) and $r$ is the rest of 'so-far-made-number' when dividing by $3$ (equivalently: $r$ is the sum of all made moves modulo $3$), and also $r\in\{0,1,2\}$. Above observation means that there are only $9$ possible positions. After choosing the $2014$-th digit by Bob we are in one of the $9$ positions: $$(0,0),\,\,(0,1),\,\,(0,2),\,\,(1,0),\,\,\ldots,\,\,(2,2).$$We will show how Alice could play (symbol of arrow) to not loose. $(0,0)\to(2,2)$, $(0,1)\to (1,2)$, $(0,2)\to(2,1)$, $(1,0)\to(2,2)$, $(1,1)\to(0,1)$, $(1,2)\to(0,2)$, $(2,0)\to(1,1)$, $(2,1)\to (0,1)$, $(2,2)\to(0,2)$. Hence Bob has to make his move ($2016$-th digit) from one of the six position: $(m,r)$ for any $m$ and $r\neq 0$. $(0,1)\to \{(1,2)\,\, \text{or}\,\,(2,0)\}$, $(0,2)\to\{(1,0)\,\,\text{or}\,\,(2,1)\}$, $(1,1)\to\{(0,1)\,\,\text{or}\,\,(2,0)\}$, $(1,2)\to \{(0,2)\,\, \text{or}\,\,(2,1)\}$, $(2,1)\to\{(0,1)\,\,\text{or}\,\,(1,2)\}$, $(2,2)\to\{(0,2)\,\,\text{or}\,\,(1,0)\}$. Now Alice's move should be: $(0,0)$ from positions $(1,0)$ or $(2,0)$ $(1,2)$ from positions $(0,1)$ or $(2,1)$ $(2,1)$ from positions $(0,2)$ or $(1,2)$. As can be easly checked Bob is not able to finish with $(m,0)$.