Solution:

The answer is no. Suppose that each of the totals from $2$ to $12$ has an equal probability, which must be $1/11$ since the sum of all probabilities is $1$. Let $a$ and $b$ be the probabilities of a $1$ and a $6$, respectively, on the first die, and let $c$ and $d$ be the corresponding probabilities on the second die.

Since $1/11$ is the probability of rolling a total of $2$, $a c = 1/11$ so $c = 1/(11a)$; since $1/11$ is the probability of rolling $12$, $b d = 1/11$ so $d = 1/(11b)$. Since the probability of rolling a $7$ through the combination $1+6$ or $6+1$ is at most $1/11$,
$$\begin{aligned} \frac{1}{11} &\geq a d + b c \\ \frac{1}{11} &\geq \frac{a}{11b} + \frac{b}{11a} \\ 1 &\geq \frac{a}{b} + \frac{b}{a} \end{aligned}$$
Since $a/b$ and $b/a$ are reciprocals, one of them is at least $1$, so this inequality cannot hold.