Distinguishable bosons
The short answer is: yes, adding a $\vec{J}1.\vec{J}_2$ term will lift some of the degeneracies from the spectrum. To see this mathematically, let us first agree to drop the parentheses on the $\vec{J}{(i)}$, and represent them as $\vec{J}i$ instead. With $$ H_0 = \vec{J}_1^2 + \vec{J}_2^2 $$ $H_0$ is diagonal in $|j_1,m_1\rangle \otimes |j_2,m_2\rangle$. From this we immediately write the eigenvalues as $\hbar^2[j_1(j+1)+j_2(j_2+1)]$. That's a total of $(2j_1+1)(2j_2+1)$ degenerate states for a given pair of $j_1$ and $j_2$. With $\vec{J}_1.\vec{J}_2$ added, the situation get a little better -- the Hamiltonian takes the form $$ H = a \vec{J}^2 + b (\vec{J}_1^2 + \vec{J}_2^2) $$ This time, the Hamiltonian is diagonal in the ''coupled'' $|j_1,j_2;jm\rangle$ basis, with the eigenvalues being $$ \hbar^2 [a j (j+1) + bj_1(j_1+1) +b j_2(j_2+1)] $$ where from the rules of addition of angular momenta, we have $$ |j_1 - j_2| \leq j \leq j_1+j_2 $$ The presence of a $j$-dependent term in the energies breaks some of the degeneracies. From the earlier $(2j_1+1)(2j_2+1)$ degeneracies, we're down to $(2j+1)$ degenerate states for each $j$. As a bonus, we can verify the counting by testing that that $$ \sum{j=|j_1-j_2|}^{j_1+j_2} (2j+1)= (2j_1+1)(2j_2+1) $$ is indeed satisfied.
Identical bosons
With this understanding, now we can specialize to the more restricted case of identical bosons. 66
Here, for $j_1=j_2$, we can seen in the $|j_1,m_1\rangle \otimes |j_2,m_2\rangle$ basis that the no. of allowed states is $$ n = \frac{1}{2}(2j_1+1)[(2j_1+1)-1] + (2j_1+1) $$ The first term counts the no. of symmetric states with $m_1 \neq m_2$, while the 2nd counts the states with $m_1=m_2$. All of these $n$ states are degenerate under the Hamiltonian $H_0$.
Now, adding the $\vec{J}_1.\vec{J}_2$ term still breaks the degeneracies, but from the smaller set of correspondingly restricted total angular momentum states. This time, the requirement of symmetrization amounts to restricting the total angular momenta to even $j$'s. So you still have the removal of degeneracies.
Example
Taking the example from your comment, for $j_1 = j_2 = 1$, we count $6$ unique, symmetrized states in the $|j_1,m_1\rangle \otimes |j_2,m_2\rangle$ basis. In the total angular momentum basis, we'd have $j \in {0,1,2}$ for distinguishable particles, but symmetrization for our identical bosons restricts us to $j \in {0,2}$. So again, instead of $6$ degenerate states, the new Hamiltonian $H$ will have $2$ separate degenerate blocks consisting of $2\times 0 +1=1$ and $2\times 2 + 1=5 $ states respectively. 976