Yes, there is a way to guess a number asking **14** questions in worst case. To do it you
need a linear code with length 14, dimension 10 and distance at least 3. One such code can be built
based on Hamming code (see http://en.wikipedia.org/wiki/Hamming_code).

Here is the strategy.

Let us denote bits of first player's number as $a_i$, $i \in [1..10]$.
We start with asking values of all those bits. That is we ask the following questions: "is it true that i-th bit of your number is zero?"
Let us denote answers on those questions as $b_i$, $i \in [1..10]$.

Now we ask **4** additional questions:

Is it true that $a_{1} \otimes a_{2} \otimes a_{4} \otimes a_{5} \otimes a_{7} \otimes a_{9}$ is equal to zero? ($\otimes$ is sumation modulo $2$).

Is it true that $a_{1} \otimes a_{3} \otimes a_{4} \otimes a_{6} \otimes a_{7} \otimes a_{10}$ is equal to zero?

Is it true that $a_{2} \otimes a_{3} \otimes a_{4} \otimes a_{8} \otimes a_{9} \otimes a_{10}$ is equal to zero?

Is it true that $a_{5} \otimes a_{6} \otimes a_{7} \otimes a_{8} \otimes a_{9} \otimes a_{10}$ is equal to zero?

Let $q_1$, $q_2$, $q_3$ and $q_4$ be answers on those additional questions. Now second player calculates $t_{i}$ ($i \in [1..4]$) --- answers on those questions based on bits $b_j$ which he previously got from first player.

Now there are 16 ways how bits $q_i$ can differ from $t_i$. Let $d_i = q_i \otimes t_i$ (hence $d_i = 1$ iff $q_i \ne t_i$).

Let us make table of all possible errors and corresponding values of $d_i$:

position of error -> $(d_1, d_2, d_3, d_4)$

no error -> (0, 0, 0, 0)

error in $b_1$ -> (1, 1, 0, 0)

error in $b_2$ -> (1, 0, 1, 0)

error in $b_3$ -> (0, 1, 1, 0)

error in $b_4$ -> (1, 1, 1, 0)

error in $b_5$ -> (1, 0, 0, 1)

error in $b_6$ -> (0, 1, 0, 1)

error in $b_7$ -> (1, 1, 0, 1)

error in $b_8$ -> (0, 0, 1, 1)

error in $b_9$ -> (1, 0, 1, 1)

error in $b_{10}$ -> (0, 1, 1, 1)

error in $q_1$ -> (1, 0, 0, 0)

error in $q_2$ -> (0, 1, 0, 0)

error in $q_3$ -> (0, 0, 1, 0)

error in $q_4$ -> (0, 0, 0, 1)

All the values of $(d_1, d_2, d_3, d_4)$ are different. Hence we can find where were an error and hence find all $a_i$.